MathematicsClass 11CBSE

Picking a Team of 3 From 10 Gives Six Times Fewer Ways Than Lining Them Up

Derive nCr and its link with nPr, use the properties of nCr to simplify and solve equations, count committees and teams under conditions, and decide when a problem needs permutations, combinations or both.

How is choosing different from arranging?

Choosing Asha, Bala and Chitra for a team is one selection, whatever order their names are written in. But making them captain, vice-captain and wicket-keeper gives **

3! = 6

different arrangements of the same three people.

A combination is a selection where order does not matter; a permutation is an arrangement where it does.**

This part covers the combination formula, its properties, selection problems with conditions, and problems that need both ideas.

How do you derive nCr = n!/(r!(n - r)!) and relate it to nPr?

**Each selection of

r

objects can be arranged in

r!

ways, so

^{n} P_{r} =^{n} C_{r} \times r!

, which gives

^{n} C_{r} = \frac{n !}{r ! ( n - r )!}

.

Worked example 1.** From

10

players:

^{10} P_{3} = 10 \times 9 \times 8 = 720^{10} C_{3} = \frac{720}{3 !} = 120

Six times fewer selections than arrangements, because each team of

3

can be lined up in

6

ways.

Worked example 2 — handshakes. Five people each shake hands once with every other:

^{5} C_{2} = \frac{5 \times 4}{2} = 10 handshakes

Worked example 3.

^{8} C_{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56

.

An everyday example. **Choosing

2

toppings from

5

for a pizza** gives

^{5} C_{2} = 10

choices, since onion-and-corn is the same as corn-and-onion.

The substance. **

^{n} C_{0} =^{n} C_{n} = 1

**: there is exactly one way to choose nothing, and one way to choose everything.

How do you use nCr = nC(n - r) and nCr + nC(r - 1) = (n + 1)Cr?

**Choosing

r

objects to take is the same as choosing

n - r

to leave, so

^{n} C_{r} =^{n} C_{n - r}

; and counting selections that include or exclude one particular object gives

^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}

.

Worked example 1.**

^{50} C_{48} =^{50} C_{2} = \frac{50 \times 49}{2} = 1225

Worked example 2 — Pascal's rule.

^{8} C_{3} +^{8} C_{2} = 56 + 28 = 84 =^{9} C_{3}

Worked example 3. If

^{n} C_{9} =^{n} C_{7}

, then

9 + 7 = n

, so

n = 16

and

^{16} C_{2} = 120

.

Worked example 4. If

^{2 n} C_{3} :^{n} C_{3} = 11 : 1

\frac{2 n ( 2 n - 1 ) ( 2 n - 2 )}{n ( n - 1 ) ( n - 2 )} = 11 \frac{4 ( 2 n - 1 )}{n - 2} = 118 n - 4 = 11 n - 22 n = 6

An everyday example. **Choosing

48

50

students to go on a trip** is the same as choosing the

2

who stay back:

1225

ways either way.

The substance. In Pascal's rule, a new object is either in the selection or not — the two cases add up to the whole count.

How do you solve selection problems on committees and teams with conditions?

Split the selection into groups according to the condition, use combinations within each group, multiply the choices, and add the separate cases — using the complement when at least is easier to count the other way.

Worked example 1. A committee of

3

from

2

men and

3

women with exactly

1

man and

2

women:

^{2} C_{1} \times^{3} C_{2} = 2 \times 3 = 6

Worked example 2 — at least one of each. A team of

5

from

6

boys and

4

girls:

^{10} C_{5} -^{6} C_{5} = 252 - 6 = 246

(an all-girl team is impossible, since there are only

4

girls).

**Worked example 3 — at least

3

girls.**

^{4} C_{3} \cdot^{6} C_{2} +^{4} C_{4} \cdot^{6} C_{1} = 4 \times 15 + 1 \times 6 = 66

Worked example 4 — fixed players. An XI from

15

players:

- A particular player always included:

^{14} C_{10} = 1001

- A particular player always excluded:

^{14} C_{11} = 364

An everyday example. Selectors picking a school cricket XI with a fixed captain face exactly the third case.

The trap. For at least one boy and one girl, subtract the impossible cases rather than adding many separate ones.

How do you decide between permutations and combinations and solve problems that need both?

Ask whether swapping two chosen objects gives a different outcome: if yes, use permutations; if no, use combinations; when a problem selects and then arranges, multiply a combination by an arrangement count.

Worked example 1 — posts versus representatives. From

30

students:

- President and secretary (different posts):

^{30} P_{2} = 870

- Two class representatives (same role):

^{30} C_{2} = 435

Worked example 2 — select then arrange. Words of

2

vowels and

3

consonants from INVOLUTE, which has vowels

I, O, U, E

and consonants

N, V, L, T

^{4} C_{2} \times^{4} C_{3} \times 5! = 6 \times 4 \times 120 = 2880

Worked example 3 — geometry. With

10

points, no three collinear:

lines =^{10} C_{2} = 45 triangles =^{10} C_{3} = 120

An everyday example. A lucky draw with three identical prizes counts combinations of winners, while one with first, second and third prizes counts permutations.

The substance. **A triangle is a selection of

3

points**, since the same three points give the same triangle in any order.

Exam tip

What earns full marks on combinations?

Say in words whether order matters, write each case separately, and show the combination symbols before evaluating.

- **

^{n} C_{r} = \frac{n !}{r ! ( n - r )!}

and

^{n} P_{r} =^{n} C_{r} \times r!

- Use

^{n} C_{r} =^{n} C_{n - r}

to shorten calculations
- Exactly conditions: multiply group choices
- At least conditions: add cases or subtract from the total
- Select then arrange: multiply by the number of arrangements

The trap.** Counting selections of

2

representatives as

^{30} P_{2}

. Swapping the two representatives gives the same pair, so use

^{30} C_{2}

Did you know

How unlikely is it to win a lottery that picks 6 numbers from 49?

A lottery draws

6

different numbers from

1

49

, in any order. The number of possible tickets is

^{49} C_{6} = \frac{49 \times 48 \times 47 \times 46 \times 45 \times 44}{6 !} = 13983816

So one ticket has a chance of about **

1

14

million.** If order mattered, the count would be

6! = 720

times larger — which shows how much difference the choice between combinations and permutations makes.

Exam relevance

How are combinations tested in JEE Main and JEE Advanced?

Combinations are central to Permutations and Combinations in JEE Main and JEE Advanced, and they reappear as the coefficients of the Binomial Theorem and in counting outcomes in Probability.

What gets asked. Selections with restrictions, dividing objects into groups, counting lines, triangles and diagonals (

^{n} C_{2} - n

for an

n

-sided polygon), and identities built on Pascal's rule.

Question types. Numerical-value and multiple-choice questions, frequently combining selection with arrangement.

The trap that costs marks. Double counting in at least problems, where the same selection is counted in more than one case.

Key takeaways

What must you be able to do from this part?

- **

^{n} C_{r} = \frac{n !}{r ! ( n - r )!}

**;

^{n} P_{r} =^{n} C_{r} \times r!

;

^{10} C_{3} = 120

- **

^{n} C_{r} =^{n} C_{n - r}

**;

^{50} C_{48} = 1225

- Pascal's rule:

^{8} C_{3} +^{8} C_{2} =^{9} C_{3} = 84

- Conditions: exactly means multiply groups; at least means add cases or subtract
- **Team of

5

with at least one boy and one girl**:

246

- Order matters: permutation; order does not matter: combination; select then arrange: multiply

Work out how many different handshakes happen when everyone in your class greets everyone else once.