# Exercise : 4

1. All possible two factors products are formed from the numbers 1, 2, 3, 4, ....., 200. The number of factors out of total obtained which are multiples of 5 is
(a) 5040
(b) 7180
(c) 8150
(d) 7280
(e) None of these

#### View Ans & Explanation

Ans.b

The total number of two factor products = 200C2. The number of numbers from 1 to 200 which are not multiples of 5 is 160. Therefore the total number of two factor products which are not multiple of 5 is 160C2. Hence,the required number of factors which are multiples of 5 = 200C2160C2 = 7180.

Directions (Qs. 2 - 3): Answer these questions on the basis of the information given below:

From a group of 6 men and 4 women a committee of 4 persons is to be formed.

2. In how many different ways can it be done so that the committee has at least one woman?
(a) 210
(b) 225
(c) 195
(d) 185
(e) None of these

#### View Ans & Explanation

Ans.c

Reqd. no. of ways

= 4C1 × 6C3 + 4C2 × 6C2 + 4C3 × 6C1 + 4C4

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

= 80 + 90 + 24 + 1 = 195

3. In how many different ways can it be done so that the committee has at least 2 men?
(a) 210
(b) 225
(c) 195
(d) 185
(e) None of these

#### View Ans & Explanation

Ans.d

Reqd. no. of ways

= 6C2 × 4C2 + 6C3 × 4C1 + 6C4

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

= 90 + 80 + 15 = 185.

4. In how many different ways can the letters of the word ORGANISE be arranged in such a way that all the vowels always come together and all the consonants always come together?
(a) 576
(b) 1152
(c) 2880
(d) 1440
(e) None of these

#### View Ans & Explanation

Ans.b

The word ORGANISE has 4 vowels and 4 consonants.

Now, both groups (vowels and consonants) can be treated as two letters. This can be arranged in 2! ways.

Now, the 4 letters of each group can be arranged.in 4! ways.

So, total possible ways of arrangement

= 2! × 4! × 4!

= 2 × 24 × 24 = 1152.