Question

(i) Find the number of ways of selecting 5 members from a committee of 5 men & 2 women such that all women are always included. (ii) Out of first 20 natural numbers, 3 numbers are selected such that there is exactly one even number. How many different selections can be made ? (iii) How many four letter words can be made from the letters of the word 'PROBLEM'. How many of these start as well as end with a vowel ?

Answer 1

(i) 10 (ii) 450 (iii) Total four-letter words: 840. Words starting and ending with a vowel: 40.

Answer 2

Here are the solutions to the given problems:

(i) Number of ways of selecting 5 members from a committee of 5 men & 2 women such that all women are always included.

To include both women, we must select the 2 women from the 2 available women. The number of ways to do this is $\binom{2}{2}$.

Since we need to select a total of 5 members and 2 women are already selected, we need to select $5 - 2 = 3$ more members. These members must be selected from the 5 men. The number of ways to do this is $\binom{5}{3}$.

The total number of ways to select the committee is the product of the number of ways to select the women and the number of ways to select the men.

Total ways = $\binom{2}{2} \times \binom{5}{3} = 1 \times \frac{5!}{3!2!} = 1 \times \frac{5 \times 4}{2 \times 1} = 1 \times 10 = 10$.

(ii) Out of first 20 natural numbers, 3 numbers are selected such that there is exactly one even number.

The first 20 natural numbers are $\{1, 2, \dots, 20\}$.

There are 10 even numbers $\{2, 4, \dots, 20\}$ and 10 odd numbers $\{1, 3, \dots, 19\}$.

We need to select 3 numbers with exactly one even number. This means we must select 1 even number and 2 odd numbers.

Number of ways to select 1 even number from 10 even numbers = $\binom{10}{1} = 10$.

Number of ways to select 2 odd numbers from 10 odd numbers = $\binom{10}{2} = \frac{10!}{2!8!} = \frac{10 \times 9}{2 \times 1} = 45$.

The total number of different selections is the product of the number of ways to select the even number and the number of ways to select the odd numbers.

Total selections = $\binom{10}{1} \times \binom{10}{2} = 10 \times 45 = 450$.

(iii) How many four letter words can be made from the letters of the word 'PROBLEM'. How many of these start as well as end with a vowel ?

The word 'PROBLEM' has 7 distinct letters: P, R, O, B, L, E, M.

Vowels: O, E (2). Consonants: P, R, B, L, M (5).

Part 1: How many four letter words can be made from the letters of the word 'PROBLEM'.

This is a permutation problem. We need to arrange 4 letters out of 7 distinct letters.

The number of four-letter words = $P(7, 4) = \frac{7!}{(7-4)!} = \frac{7!}{3!} = 7 \times 6 \times 5 \times 4 = 840$.

Part 2: How many of these four-letter words start as well as end with a vowel.

Let the four-letter word be represented by 4 positions: $L_1 L_2 L_3 L_4$.

$L_1$ must be a vowel, and $L_4$ must be a vowel. There are 2 vowels (O, E).

Number of ways to choose and arrange 2 vowels for positions $L_1$ and $L_4$ is $P(2, 2) = 2! = 2$. (The vowels can be O at $L_1$ and E at $L_4$, or E at $L_1$ and O at $L_4$).

The remaining $7 - 2 = 5$ letters are available to fill the middle two positions ($L_2$ and $L_3$). These 5 letters are the consonants.

We need to choose and arrange 2 letters from these 5 remaining letters for positions $L_2$ and $L_3$. The number of ways to do this is $P(5, 2) = \frac{5!}{(5-2)!} = \frac{5!}{3!} = 5 \times 4 = 20$.

The total number of four-letter words that start and end with a vowel is the product of the number of ways to fill the end positions and the number of ways to fill the middle positions.

Total words = (Ways to fill $L_1$ and $L_4$) $\times$ (Ways to fill $L_2$ and $L_3$) = $P(2, 2) \times P(5, 2) = 2 \times 20 = 40$.

Explanation (Minimal):

(i) Select 2 women from 2 ($\binom{2}{2}$) and 3 men from 5 ($\binom{5}{3}$). Multiply the results: $\binom{2}{2} \times \binom{5}{3} = 1 \times 10 = 10$.

(ii) Select 1 even number from 10 ($\binom{10}{1}$) and 2 odd numbers from 10 ($\binom{10}{2}$). Multiply the results: $\binom{10}{1} \times \binom{10}{2} = 10 \times 45 = 450$.

(iii) Total 4-letter words: Permute 4 out of 7 distinct letters ($P(7, 4) = 840$). Words starting and ending with vowel: Arrange 2 vowels at ends ($P(2, 2)$) and arrange 2 from remaining 5 letters in middle ($P(5, 2)$). Multiply results: $P(2, 2) \times P(5, 2) = 2 \times 20 = 40$.