Chapter 6 · Permutations and Combinations

6.3 Combinations

When only the choice matters and not the order, every selection of $r$ objects has been counted $r!$ times among the permutations — so ${}^nC_r=\dfrac{{}^nP_r}{r!}=\dfrac{n!}{r!\,(n-r)!}$.

A selection committee has a squad of $15$ cricketers and must pick a playing eleven. Unlike the batting order of Section 6.2, the eleven is just a group: picking Arjun, then Dev, then Kabir gives the same team as picking Kabir, then Arjun, then Dev. Permutations count those as different, so ${}^{15}P_{11}$ would overcount — but by exactly how much? Each team of $11$ can be put in order in $11!$ ways, so every team appears $11!$ times among the permutations. Dividing removes the overcount: there are $\dfrac{{}^{15}P_{11}}{11!}=1365$ possible teams.

A selection in which order does not matter is called a combination. In this section we prove the formula for the number of combinations ${}^nC_r$, establish its two most important properties — the symmetry ${}^nC_r={}^nC_{n-r}$ and the addition rule ${}^nC_r+{}^nC_{r-1}={}^{n+1}C_r$ — and then solve the mixed problems that dominate exams: committees with conditions, and "select and then arrange" problems that need both ${}^nC_r$ and ${}^nP_r$.

Selections and the Formula for ${}^nC_r$

📘 Definition 6.4 — Combination

A combination is a selection of some or all of a collection of objects, without regard to order. For integers $0\le r\le n$, the number of combinations of $n$ different objects taken $r$ at a time is written $${}^nC_r\quad\text{or}\quad\binom{n}{r}.$$ It is the number of subsets with $r$ elements of a set with $n$ elements. For example, the $2$-element selections from $\{\text{A},\text{B},\text{C},\text{D}\}$ are $\text{AB},\text{AC},\text{AD},\text{BC},\text{BD},\text{CD}$, so ${}^4C_2=6$.

Compare with the permutations of the same letters taken $2$ at a time: $\text{AB},\text{BA},\text{AC},\text{CA},\dots$ — there are ${}^4P_2=12$, exactly $2!=2$ for each selection. The next theorem turns this observation into a formula.

📐 Theorem 6.7 — Formula for ${}^nC_r$

For integers $0\le r\le n$, $${}^nP_r={}^nC_r\times r!,$$ and therefore $${}^nC_r=\dfrac{n!}{r!\,(n-r)!}.$$ In particular ${}^nC_0={}^nC_n=1$ and ${}^nC_1=n$.

Proof. Arranging $r$ of the $n$ objects in a row can be done in two stages: first select which $r$ objects to use (${}^nC_r$ ways), then arrange the $r$ selected objects in order ($r!$ ways, by Theorem 6.4 with $n=r$). The number of arrangements at the second stage does not depend on which objects were selected, so by the fundamental principle of counting the number of arrangements is ${}^nC_r\times r!$. But that number is ${}^nP_r$. Hence ${}^nC_r\cdot r!={}^nP_r=\dfrac{n!}{(n-r)!}$, and dividing by $r!$ gives the formula. For $r=0$ or $r=n$ it gives $\dfrac{n!}{0!\,n!}=1$ — one way to choose nothing, one way to choose everything. $\blacksquare$

✏️ Example 1 — Evaluating ${}^nC_r$
Evaluate (a) ${}^8C_3$ (b) ${}^{10}C_8$ (c) ${}^{15}C_{13}$ (d) ${}^{15}C_{11}$.
(a) $\dfrac{8!}{3!\,5!}=\dfrac{8\times7\times6}{3\times2\times1}=56$. (Cancel $5!$ first: the top keeps $r$ factors counting down from $n$, the bottom is $r!$.)
(b) $\dfrac{10!}{8!\,2!}=\dfrac{10\times9}{2}=45$.
(c) $\dfrac{15!}{13!\,2!}=\dfrac{15\times14}{2}=105$.
(d) $\dfrac{15!}{11!\,4!}=\dfrac{15\times14\times13\times12}{24}=1365$ — the number of playing elevens in the introduction. Notice that (b), (c) and (d) were easiest with the smaller of $r$ and $n-r$ on the bottom; Theorem 6.8 explains why that is always allowed.
✏️ Example 2 — Committees, handshakes and triangles
(a) In how many ways can a committee of $3$ be chosen from $8$ people?
(b) $12$ people at a meeting each shake hands once with every other person. How many handshakes are there?
(c) $10$ points are marked on a circle. How many straight lines and how many triangles have their vertices among these points?
(a) A committee is a selection: ${}^8C_3=56$.
(b) A handshake is an unordered pair of people: ${}^{12}C_2=\dfrac{12\times11}{2}=66$.
(c) No three points on a circle are collinear, so every pair of points gives a different line and every triple a different triangle: lines ${}^{10}C_2=45$, triangles ${}^{10}C_3=120$.
Check (b) Each of the $12$ people shakes $11$ hands, giving $12\times11=132$ — but each handshake involves two people and is counted twice: $\tfrac{132}{2}=66$ ✓. This "divide out the double count" is exactly the $r!$ of Theorem 6.7 with $r=2$.
Interactive 1 — Permutations Are Combinations Times $r!$

Each row on the left is one selection (a combination) of $r$ letters from the first $n$ letters. The highlighted selection fans out on the right into all its orderings — always $r!$ of them. So the permutations come in ${}^nC_r$ groups of $r!$: ${}^nP_r={}^nC_r\times r!$.

selections ${}^nC_r$—
orderings of each, $r!$—
permutations ${}^nP_r$—
groups seen so far—

Properties of ${}^nC_r$

📐 Theorem 6.8 — Choosing is the same as leaving out

For integers $0\le r\le n$, $${}^nC_r={}^nC_{n-r}.$$ Consequently, if ${}^nC_a={}^nC_b$ (with $0\le a,b\le n$), then $a=b$ or $a+b=n$.

Proof (algebra). ${}^nC_{n-r}=\dfrac{n!}{(n-r)!\,\big(n-(n-r)\big)!}=\dfrac{n!}{(n-r)!\,r!}={}^nC_r$.

Proof (counting). Choosing the $r$ objects you take is the same as choosing the $n-r$ objects you leave behind: each selection of $r$ matches exactly one selection of $n-r$ (its complement), so the two numbers are equal.

The consequence. As $r$ goes from $0$ up to $\tfrac n2$, the values ${}^nC_r$ strictly increase, because $\dfrac{{}^nC_{r}}{{}^nC_{r-1}}=\dfrac{n-r+1}{r}\gt1$ exactly when $r\lt\tfrac{n+1}{2}$; after the middle they decrease symmetrically. So each value occurs at most twice in the row ${}^nC_0,\dots,{}^nC_n$, at $r$ and at $n-r$. Hence ${}^nC_a={}^nC_b$ forces $b=a$ or $b=n-a$. $\blacksquare$

📐 Theorem 6.9 — The addition rule ${}^nC_r+{}^nC_{r-1}={}^{n+1}C_r$

For integers $1\le r\le n$, $${}^nC_r+{}^nC_{r-1}={}^{n+1}C_r.$$

Proof (algebra). Write both terms over the common denominator $r!\,(n-r+1)!$, using $(n-r+1)!=(n-r+1)(n-r)!$ and $r!=r\,(r-1)!$: $$\begin{aligned}{}^nC_r+{}^nC_{r-1}&=\dfrac{n!\,(n-r+1)}{r!\,(n-r+1)!}+\dfrac{n!\;r}{r!\,(n-r+1)!}\\&=\dfrac{n!\,(n+1)}{r!\,(n+1-r)!}=\dfrac{(n+1)!}{r!\,(n+1-r)!}={}^{n+1}C_r.\end{aligned}$$

Proof (counting). Choose $r$ objects from $n+1$ objects, one of which is special — call it S. Every selection either leaves S out — then all $r$ objects come from the other $n$: ${}^nC_r$ ways — or includes S — then the other $r-1$ come from the other $n$: ${}^nC_{r-1}$ ways. The two cases do not overlap, so by the addition principle ${}^{n+1}C_r={}^nC_r+{}^nC_{r-1}$. $\blacksquare$

The counting proof is worth remembering for its own sake: "split on whether a particular object is chosen" solves many problems in which one person must, or must not, be on a committee. Theorem 6.9 is also the rule that builds the triangle of numbers in Chapter 7 (Binomial Theorem), where each entry is the sum of the two above it.

✏️ Example 3 — Using the properties
(a) If ${}^nC_9={}^nC_8$, find ${}^nC_{17}$.
(b) If ${}^{2n}C_3:{}^nC_3=11:1$, find $n$.
(c) Evaluate ${}^7C_3+{}^7C_2$ in one step.
(a) $9\ne8$, so by Theorem 6.8, $9+8=n$: $n=17$, and ${}^{17}C_{17}=1$.
(b) Set up $\dfrac{2n(2n-1)(2n-2)}{n(n-1)(n-2)}=11$. Cancel $2n-2=2(n-1)$ and $n$: $\dfrac{4(2n-1)}{n-2}=11$, so $8n-4=11n-22$ and $n=6$.
(b) Check ${}^{12}C_3=220$ and ${}^6C_3=20$; $220:20=11:1$ ✓.
(c) Theorem 6.9 with $n=7$, $r=3$: ${}^7C_3+{}^7C_2={}^8C_3=56$. Check: $35+21=56$ ✓.
Interactive 2 — The ${}^nC_r$ Table and the Addition Rule

Row $n$ of the table lists ${}^nC_0,{}^nC_1,\dots,{}^nC_n$. Pick a cell: its two neighbours in the row above, ${}^{n-1}C_{r-1}$ and ${}^{n-1}C_r$, are highlighted in red and blue, and they add up to it (Theorem 6.9). The green cell is its mirror image ${}^nC_{n-r}$ — always equal (Theorem 6.8). Press Play to walk through the table.

62
${}^{n-1}C_{r-1}$—
${}^{n-1}C_r$—
sum $={}^nC_r$—
mirror ${}^nC_{n-r}$—

Mixed Problems: Select, Then Arrange

🧭 Problem-solving strategy — Combination or permutation?
  1. Ask: does order matter? Swap two chosen objects. If the outcome is different (a password, a batting order, president vs secretary), order matters — permutation. If it is the same (a committee, a team, a hand of cards), it does not — combination.
  2. Select first, with conditions. Split the selection into groups (men/women, batters/bowlers) and multiply the ${}^nC_r$ of the groups. For "at least" or "at most" conditions, list the cases and add, or subtract the unwanted cases from the total.
  3. Then arrange, if needed. If the chosen objects are to be put in order, multiply by $r!$ (or by the alike-objects formula of Section 6.2).
  4. Never "choose one first, then choose the rest" for an "at least one" condition — it counts some selections more than once (see the warning below).
✏️ Example 4 — A committee with conditions
A committee of $5$ is to be chosen from $6$ men and $4$ women. In how many ways can this be done if the committee has (a) exactly $2$ women (b) at least $2$ women (c) a particular man M and a particular woman W are not both on it?
(a) $2$ of the $4$ women and $3$ of the $6$ men: ${}^4C_2\times{}^6C_3=6\times20=120$.
(b) Cases $2$ women: $120$. $3$ women: ${}^4C_3\times{}^6C_2=4\times15=60$. $4$ women: ${}^4C_4\times{}^6C_1=6$. Total $120+60+6=186$.
(b) Check Subtract instead: all committees ${}^{10}C_5=252$; with $0$ women ${}^6C_5=6$; with $1$ woman ${}^4C_1\times{}^6C_4=4\times15=60$. $252-6-60=186$ ✓.
(c) Committees containing both M and W: the other $3$ members from the remaining $8$: ${}^8C_3=56$. Not both: $252-56=196$.
✏️ Example 5 — Select, then arrange
From $5$ different vowels and $7$ different consonants, how many words (with or without meaning) of $5$ different letters can be made that contain exactly $2$ vowels and $3$ consonants?
Select Vowels ${}^5C_2=10$; consonants ${}^7C_3=35$. Selections of letters: $10\times35=350$.
Arrange Each selection of $5$ different letters can be arranged in $5!=120$ orders.
Answer $350\times120=42\,000$ words.
Check Equivalently: choose which $2$ of the $5$ positions hold vowels (${}^5C_2=10$), fill them in order (${}^5P_2=20$), and fill the other $3$ positions with consonants in order (${}^7P_3=210$): $10\times20\times210=42\,000$ ✓.
✏️ Example 6 — Picking a playing eleven
An IPL squad of $18$ has $7$ batters, $6$ bowlers, $3$ all-rounders and $2$ wicketkeepers. In how many ways can an eleven be chosen with exactly $1$ wicketkeeper, $4$ batters, $4$ bowlers and $2$ all-rounders? If, in addition, a captain and a vice-captain are named from the eleven, how many choices are there?
Select by group ${}^2C_1\times{}^7C_4\times{}^6C_4\times{}^3C_2=2\times35\times15\times3=3150$ elevens. (Check the total: $1+4+4+2=11$ ✓.)
Leaders Captain and vice-captain are different jobs, so order matters: ${}^{11}P_2=110$ for each eleven.
Answer $3150\times110=346\,500$ choices of team with captain and vice-captain.
⚠️ Common mistakes with combinations

1. Using ${}^nP_r$ for a selection. A committee of $3$ from $8$ is ${}^8C_3=56$, not ${}^8P_3=336$. Swap two members — if nothing changes, use ${}^nC_r$.

2. "Choose one, then the rest" overcounts. Committees of $5$ from $6$ men and $4$ women with at least one woman: "choose $1$ woman (${}^4C_1$), then any $4$ of the other $9$ (${}^9C_4$)" gives $4\times126=504$ — more than the $252$ committees that exist! A committee with two women W₁, W₂ is counted once with W₁ "chosen first" and again with W₂. The correct count is ${}^{10}C_5-{}^6C_5=252-6=246$.

3. Forgetting to arrange after selecting. "How many words…" needs ${}^nC_r\times r!$; "how many selections…" does not.

4. Misusing the symmetry. ${}^nC_a={}^nC_b$ gives $a=b$ or $a+b=n$ — not just $a+b=n$. And ${}^nC_r$ needs $0\le r\le n$: ${}^5C_7$ is not defined here (there is no way to choose $7$ from $5$).

5. Points on a line. If some of the points are collinear, ${}^nC_2$ overcounts the lines and ${}^nC_3$ counts "triangles" that are flat. Subtract the collinear cases and add back the one line through them.

🚀 Beyond the board syllabus (JEE Main · JEE Advanced)

In JEE Main and JEE Advanced 2026 (both list permutations, combinations and their applications); not in the CBSE year-end exam. Both regularly ask how to distribute objects into groups or boxes.

Distinct objects into groups. $m+n$ different objects can be split into a group of $m$ and a group of $n$ (with $m\ne n$) in ${}^{m+n}C_m=\dfrac{(m+n)!}{m!\,n!}$ ways. If the two groups have the same size $n$ and are not labelled, divide by $2!$: $\dfrac{(2n)!}{2!\,(n!)^2}$, because swapping the two groups gives the same split.

Identical objects into distinct boxes ("stars and bars"). The number of ways to put $n$ identical objects into $r$ different boxes (boxes may be empty) is ${}^{n+r-1}C_{r-1}$: write the objects as $n$ stars and separate the boxes by $r-1$ bars; each distribution is one arrangement of $n$ stars and $r-1$ bars, and there are $\dfrac{(n+r-1)!}{n!\,(r-1)!}$ of them (Section 6.2).

Worked example. In how many ways can $10$ identical sweets be shared among $3$ children? In how many if every child gets at least one?
Solution. $n=10$, $r=3$: ${}^{12}C_2=66$ ways. For "at least one each", first give every child one sweet; the remaining $7$ are shared freely: ${}^{9}C_2=36$ ways. Check: the number of solutions of $a+b+c=10$ in whole numbers is $\sum_{a=0}^{10}(11-a)=11+10+\cdots+1=66$ ✓.

Exercises

I
Evaluating and solving.
  • 1 Evaluate ${}^9C_4$.
  • 2 Evaluate ${}^{12}C_{10}$.
  • 3 Evaluate ${}^{20}C_{18}$ using Theorem 6.8.
  • 4 Find $n$ if ${}^nC_2=28$.
  • 5 If ${}^nC_{10}={}^nC_{12}$, find ${}^nC_{20}$.
  • 6 If ${}^{18}C_r={}^{18}C_{r+2}$, find ${}^rC_5$.
  • 7 If ${}^nP_r=840$ and ${}^nC_r=35$, find $r$ and $n$.
  • 8 Verify that ${}^9C_4+{}^9C_3={}^{10}C_4$ by evaluating both sides.
  • 9 Show that ${}^5C_1+{}^5C_2+{}^5C_3+{}^5C_4+{}^5C_5=31$, and explain the answer by counting the non-empty subsets of a $5$-element set.
1 $\dfrac{9\times8\times7\times6}{24}=126$.
2 ${}^{12}C_2=\dfrac{12\times11}{2}=66$.
3 ${}^{20}C_2=\dfrac{20\times19}{2}=190$.
4 $\dfrac{n(n-1)}{2}=28$, $n(n-1)=56$, $n=8$.
5 $10\ne12$, so $n=22$; ${}^{22}C_{20}={}^{22}C_2=231$.
6 $r\ne r+2$, so $r+(r+2)=18$, $r=8$; ${}^8C_5={}^8C_3=56$.
7 $r!=\dfrac{{}^nP_r}{{}^nC_r}=24$, so $r=4$. Then $n(n-1)(n-2)(n-3)=840=7\times6\times5\times4$, so $n=7$.
8 $126+84=210$ and ${}^{10}C_4=\dfrac{10\times9\times8\times7}{24}=210$ ✓.
9 $5+10+10+5+1=31$. Each term counts the subsets of one size; altogether they are all subsets except $\phi$. Each of the $5$ elements is in or out ($2^5=32$ subsets, by Section 6.1), and $32-1=31$ ✓.
II
Selections.
  • 1 A student must answer $4$ questions out of $7$. In how many ways can she choose them?
  • 2 In how many ways can a quiz team of $3$ be chosen from $12$ students?
  • 3 How many handshakes take place when each of $15$ people shakes hands once with each of the others?
  • 4 How many diagonals does a decagon ($10$-sided polygon) have?
  • 5 $12$ points lie in a plane, and exactly $5$ of them lie on one straight line (no other three are collinear). How many straight lines, and how many triangles, can be formed with vertices among the points?
  • 6 How many $5$-card hands consisting only of hearts can be dealt from a standard pack ($13$ hearts)?
  • 7 A committee of $4$ is chosen from $5$ boys and $6$ girls. In how many ways if it has (a) exactly $2$ boys and $2$ girls (b) at least $3$ girls?
  • 8 In how many ways can $5$ players be chosen from $10$ if (a) the captain must be included (b) a particular injured player must be left out?
Check: committees with a condition, counted by listing
by formula (sum of cases)—
by listing every committee—
1 ${}^7C_4=35$.
2 ${}^{12}C_3=220$.
3 ${}^{15}C_2=105$.
4 Pairs of vertices give ${}^{10}C_2=45$ segments; $10$ of them are sides. Diagonals: $35$.
5 Lines: ${}^{12}C_2-{}^5C_2+1=66-10+1=57$ (the $10$ pairs from the collinear points all give the same single line). Triangles: ${}^{12}C_3-{}^5C_3=220-10=210$.
6 ${}^{13}C_5=1287$.
7 (a) ${}^5C_2\times{}^6C_2=10\times15=150$. (b) $3$ girls: ${}^6C_3\times{}^5C_1=100$; $4$ girls: ${}^6C_4=15$; total $115$.
8 (a) The captain is in; choose $4$ more from $9$: ${}^9C_4=126$. (b) Choose all $5$ from the other $9$: ${}^9C_5=126$. (Equal, by Theorem 6.8.)
III
Select, then arrange.
  • 1 From $4$ different vowels and $6$ different consonants, how many words of $4$ different letters with exactly $2$ vowels and $2$ consonants can be made?
  • 2 How many $4$-letter words, each with $2$ vowels and $2$ consonants and no repeated letter, can be made from the letters of $\text{EQUATION}$?
  • 3 From a class of $20$, a president, a secretary and a $3$-member organising committee (from the remaining students) are to be chosen. In how many ways?
  • 4 In how many ways can $3$ boys and $2$ girls be chosen from $6$ boys and $4$ girls and then seated in a row?
  • 5 How many rectangles (of all sizes) are there on an $8\times8$ chessboard? (A rectangle is fixed by $2$ of the $9$ horizontal lines and $2$ of the $9$ vertical lines.)
  • 6 How many $6$-digit numbers have digits that strictly increase from left to right (like $134689$)?
  • 7 A committee of $5$ is chosen from $7$ Indians and $4$ foreigners. In how many ways if it has at least $1$ and at most $2$ foreigners?
  • 8 From $10$ people, including a married couple, $4$ are to be chosen. In how many ways if the couple are either both chosen or both left out?
1 ${}^4C_2\times{}^6C_2\times4!=6\times15\times24=2160$.
2 Vowels E, U, A, I, O ($5$); consonants Q, T, N ($3$). ${}^5C_2\times{}^3C_2\times4!=10\times3\times24=720$.
3 President and secretary are different posts, so order matters: $20\times19=380$. The committee is a selection of $3$ from the remaining $18$ students: ${}^{18}C_3=816$. Total $380\times816=310\,080$.
4 ${}^6C_3\times{}^4C_2\times5!=20\times6\times120=14\,400$.
5 ${}^9C_2\times{}^9C_2=36\times36=1296$.
6 The digit $0$ cannot appear (it would have to come first). Any $6$ of the digits $1$–$9$ can be written in increasing order in exactly one way: ${}^9C_6=84$.
7 $1$ foreigner: ${}^4C_1\times{}^7C_4=4\times35=140$; $2$ foreigners: ${}^4C_2\times{}^7C_3=6\times35=210$. Total $350$.
8 Both: choose $2$ more from the other $8$: ${}^8C_2=28$. Neither: ${}^8C_4=70$. Total $98$.
IV
Applications.
  • 1 In a lottery, $6$ numbers are drawn from $1$ to $49$. How many different tickets (sets of $6$ numbers) are possible? How many tickets match exactly $5$ of the $6$ numbers drawn?
  • 2 At an IPL auction a franchise wants $4$ overseas players from a shortlist of $9$ and $3$ uncapped Indian players from a shortlist of $7$. How many different sets of signings are possible?
  • 3 A pizza shop offers $8$ toppings. How many different pizzas have exactly $3$ toppings? How many have at most $2$ toppings (a plain pizza counts)?
  • 4 Quality control. A batch of $20$ bulbs contains $3$ defective ones. A sample of $4$ is tested. How many samples contain no defective bulb, and how many contain exactly one?
  • 5 A hockey squad of $12$ includes exactly $2$ goalkeepers. How many teams of $7$ contain exactly one goalkeeper?
  • 6 A convex polygon has $44$ diagonals. How many sides does it have?
1 ${}^{49}C_6=13\,983\,816$. Exactly $5$ correct: ${}^6C_5\times{}^{43}C_1=6\times43=258$.
2 ${}^9C_4\times{}^7C_3=126\times35=4410$.
3 ${}^8C_3=56$. At most $2$: ${}^8C_0+{}^8C_1+{}^8C_2=1+8+28=37$.
4 None defective: ${}^{17}C_4=2380$. Exactly one: ${}^3C_1\times{}^{17}C_3=3\times680=2040$.
5 ${}^2C_1\times{}^{10}C_6=2\times210=420$.
6 Diagonals $={}^nC_2-n=\dfrac{n(n-3)}{2}=44$, so $n^2-3n-88=0$, $(n-11)(n+8)=0$, $n=11$.
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE MainJEE Advanced

How this is examined: Board papers use ${}^nC_a={}^nC_b$ (1–2 marks), committee and "select then arrange" problems (3–5 marks) and case studies built on team selection. JEE favours geometry counts (lines, triangles, diagonals), "no two consecutive" selections and the distribution results of the 🚀 box.

E1. Board · 1 mark If ${}^nC_8={}^nC_6$, then ${}^nC_2$ equals (A) $91$   (B) $182$   (C) $14$   (D) $28$

Idea $8\ne6$, so $n=8+6=14$, and ${}^{14}C_2=\dfrac{14\times13}{2}=91$.
Trap (B) $=14\times13={}^{14}P_2$ forgets to divide by $2!$; (C) is $n$ itself.
Answer (A).

E2. Board · 2 marks Find $n$ if ${}^nC_{n-2}=45$.

Symmetry ${}^nC_{n-2}={}^nC_2=\dfrac{n(n-1)}{2}$. (1 mark)
Answer $n(n-1)=90=10\times9$, so $n=10$ (the root $n=-9$ is rejected). (1 mark)

E3. Board · 3 marks A bag contains $4$ red, $3$ white and $2$ blue balls, all different. In how many ways can $3$ balls be drawn so that (i) one ball of each colour is drawn (ii) at least one red ball is drawn (iii) all three are of the same colour?

(i) ${}^4C_1\times{}^3C_1\times{}^2C_1=24$. (1 mark)
(ii) All draws ${}^9C_3=84$; no red ${}^5C_3=10$; at least one red $84-10=74$. (1 mark)
Answer (iii) All red ${}^4C_3=4$ or all white ${}^3C_3=1$ (only $2$ blue balls): $4+1=5$. (1 mark)

E4. Board · Case study A franchise has a squad of $16$: $7$ batters, $5$ bowlers, $2$ all-rounders and $2$ wicketkeepers. The playing eleven must contain exactly $1$ wicketkeeper and both all-rounders; the other $8$ places go to batters and bowlers. (i) In how many ways can the wicketkeeper be chosen? (1 mark) (ii) Ignoring any further condition, in how many ways can the other $8$ places be filled? (1 mark) (iii) The coach insists on at least $4$ bowlers. How many elevens are now possible? (2 marks)

(i) ${}^2C_1=2$. (1 mark)
(ii) $8$ from the $12$ batters and bowlers: ${}^{12}C_8={}^{12}C_4=495$. (1 mark)
(iii) Cases $4$ bowlers and $4$ batters: ${}^5C_4\times{}^7C_4=5\times35=175$. $5$ bowlers and $3$ batters: ${}^5C_5\times{}^7C_3=35$. Together $210$. (1 mark)
Answer (iii) Wicketkeeper $\times$ all-rounders $\times$ the rest: $2\times{}^2C_2\times210=420$ elevens. (1 mark)

E5. JEE Main The number of triangles whose vertices are vertices of a regular $12$-gon but none of whose sides is a side of the $12$-gon is (A) $220$   (B) $112$   (C) $108$   (D) $120$

Idea Count all triangles, then subtract those using one or two sides of the polygon. All: ${}^{12}C_3=220$.
Subtract Two sides (three consecutive vertices): $12$ triangles. Exactly one side: choose the side ($12$ ways) and a third vertex not adjacent to either end ($12-4=8$ ways): $96$. So $220-12-96=112$.
Trap (C) $108=12+96$ counts the triangles that do use a side of the polygon; (A) counts all triangles. ⏱ Fast route: the known count for an $n$-gon is $\dfrac{n(n-4)(n-5)}{6}=\dfrac{12\times8\times7}{6}=112$.
Answer (B).

E6. JEE Advanced Numerical answer. In how many ways can $3$ numbers be chosen from $1,2,\dots,20$ so that no two of the chosen numbers are consecutive?

Idea Write the chosen numbers as $a\lt b\lt c$ with $b\ge a+2$ and $c\ge b+2$. Then $a\lt b-1\lt c-2$ are three different numbers from $1$ to $18$, and every such triple comes from exactly one allowed choice (add back $0,1,2$).
Count So the allowed choices match the $3$-element subsets of $\{1,\dots,18\}$: ${}^{18}C_3=\dfrac{18\times17\times16}{6}=816$.
⏱ Fast route Gap method: the $17$ numbers not chosen leave $18$ gaps; put the $3$ chosen numbers in $3$ different gaps: ${}^{18}C_3$.
Answer $816$.
🔬 Research–Teaching NexusMini project · about 45 minutes

Research question. In a fantasy-cricket game you pick $11$ players from the $22$ in a match, plus a captain and a vice-captain. How many different entries are possible — and what do the game's rules do to that number?

Why it matters. Millions of people enter such contests for a single match. Whether two entries are likely to be identical is a counting question, and the rules (limits on players from one team, required numbers of bowlers or wicketkeepers) are exactly the "conditions" of this section's committee problems.

What to do.

  1. Count the possible elevens chosen from $22$ players with no rules, and then the number of entries once a captain and a different vice-captain are named from the eleven.
  2. Add the rule "at most $7$ players from either team". Count the elevens by cases (how many come from team A), using ${}^{11}C_k\times{}^{11}C_{11-k}$.
  3. Look at a real match line-up: count the wicketkeepers, batters, all-rounders and bowlers in each team, invent a sensible rule (for example "at least $3$ bowlers"), and count again — use the check-lab idea, a spreadsheet, or a short program.
  4. Compare your counts with the number of people who might enter, and decide whether duplicate entries are likely.

Tools: A spreadsheet app (COMBIN(n, r)); optionally a short program to check by listing

Hand in: Your counts with the working for each rule, and a paragraph on how each rule changed the number of entries.

Going further. If $10$ million people chose their elevens completely at random, roughly how many pairs of identical elevens would you expect? (Hint: there are ${}^{10^7}C_2$ pairs of people.)

Instructor note: ${}^{22}C_{11}=705\,432$ elevens; with captain and vice-captain $705\,432\times110=77\,597\,520$. With "at most $7$ from either team" ($k=4$ to $7$ from team A): $644\,688$ elevens. Since $10^7$ people exceed $705\,432$, many identical elevens are certain (pigeonhole); the expected number of identical pairs is about $\tfrac{(10^7)^2/2}{705\,432}\approx7\times10^7$ — popular players make it far more.

🧪 Self-check quiz

0/5 correct

Q1. ${}^{10}C_3$ is

Q2. If ${}^nC_4={}^nC_6$, then $n$ is

Q3. ${}^7C_3+{}^7C_2$ equals

Q4. A committee of $3$ is chosen from $5$ men and $4$ women. The number of committees with exactly one woman is

Q5. In how many ways can a captain and a vice-captain be chosen from a team of $11$?

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