Basic Combinations - Permutations and Combinations

Combinations

Each of the different groups or selections which can be made by taking some or all of a number of things at a time (irrespective of order) is called a combination.

Difference between permutations and combinations.

The process of selecting things is called combination and that of arranging things is called permutations.

For example - Suppose we have 4 objects A, B, C, and D. Then,

Basic Combination Formulas

• The number of all combinations of n distinct objects, taken r at a time is given by ⁿC_r = $\frac{n!}{r!(n-r)!}\text{=}\frac{^nP_r}{r!}$
• $^{n}C_r$ + $^{n}C_{r-1}\text{ = }^{n+1}C_{r}$
• if $^{n}C_x\text{ = }^nC_y\text{ then, n = x+y }$

Practice Questions

Question 1. Determine n, if $^{2n}C_3$ : $^{n}C_2$ = 12 : 1

Solution:

$^{2n}C_3$ : $^{n}C_2 \Rightarrow$ 10 + 12 = n $\Rightarrow$ n = 22

Now, $^nC_5 \text{ = }^{22}C_5 \text{ = }\frac{22!}{5!(22-5)!} \text{ = }\frac{22!}{5!(17)!} \text{ = }\frac{22*21*20*19*18*17!}{5*4*3*2*1*17!}$ = 11 * 21 * 19 * 6 = 26334

Question 3. How many ways can a committee of 4 people be chosen from a group of 10 people?

Solution:

$^{10}C_4 = \frac{10!}{4!(10-4)!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$

Question 4. From a deck of 52 cards, how many ways can you draw 5 cards such that all of them are hearts?

Solution:

$^{13}C_5 = \frac{13!}{5!(13-5)!} = \frac{13 \times 12 \times 11 \times 10 \times 9}{5 \times 4 \times 3 \times 2 \times 1} = 1287$

Question 5. In how many ways can 5 different books be arranged on a shelf?

Solution:

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

Question 6. In an examination, a question paper consists of 10 questions. A student is required to answer any 6 questions. In how many ways can the student choose the questions?

Solution:

$^{10}C_6 = \frac{10!}{6!(10-6)!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$

Question 7. How many ways can you choose 3 balls from a box containing 6 red, 4 green, and 5 blue balls?

Solution:

Total balls = 6 + 4 + 5 = 15

$^{15}C_3 = \frac{15!}{3!(15-3)!} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455$

Question 8. From 7 men and 5 women, a committee of 4 people is to be formed. In how many ways can this be done if the committee must include at least 2 women?

Solution:

We need to consider three scenarios: 2 women and 2 men, 3 women and 1 man, 4 women

Case 1: 2 women and 2 men

$^5C_2 \times ^7C_2 = \frac{5!}{2!3!} \times \frac{7!}{2!5!} = 10 \times 21 = 210$

Case 2: 3 women and 1 man

$^5C_3 \times ^7C_1 = \frac{5!}{3!2!} \times \frac{7!}{1!6!} = 10 \times 7 = 70$

Case 3: 4 women

$^5C_4 = \frac{5!}{4!1!} = 5$

Total number of ways:

210 + 70 + 5 = 285

Question 2. In an examination, a question paper consists of 12 questions divided into two parts I and II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can students select the questions?

Solution:

If S is the required number of ways, then

S = $^5C_3 \times \text{ } ^7C_5$ + $^5C_4 \times \text{ } ^7C_4$ + $^5C_5 \times \text{ } ^7C_3$

= $\frac{5!}{3!2!} \times \frac{7!}{5!2!} + \frac{5!}{4!1!} \times \frac{7!}{4!3!} + \frac{5!}{5!0!} \times \frac{7!}{3!4!}$

= $\frac{5 \times 4 \times 3!}{3! \times 2 \times 1} \times \frac{7 \times 6 \times 5!}{5! \times 2 \times 1} + \frac{5 \times 4!}{4! \times 1} \times \frac{7 \times 6 \times 5 \times 4!}{4! \times 3 \times 2 \times 1} + \frac{5!}{5!} \times \frac{7 \times 6 \times 5 \times 4!}{3! \times 2 \times 1 \times 4!}$

= $(5 \times 2 \times 7 \times 3) + (5 \times 7 \times 5) + (7 \times 5)$

= 210 + 175 + 35

= 420 ways