Exercise : 1


1. In how many different ways can be letters of the word SOFTWARE be arranged in such a way that the vowels always come together?
(a) 13440
(b) 1440
(c) 360
(d) 120
(e) None of these
Ans.e

O,A,E S F T W R

When the vowels are always together, then treat all the vowels as a single letter and then all the letters can be arranged in 6! ways and also all three vowels can be arranged in 3! ways. Hence, required no. of arrangements = 6! × 3! = 4320.

2. In how many different ways can a group of 4 men and 4 women be formed out of 7 men and 8 women?
(a) 2450
(b) 105
(c) 1170
(d) Cannot be determined
(e) None of these
Ans.a

Reqd no. of ways = 7C4 × 8C4

= \(\frac{7 \times 6 \times 5 \times 4}{1 \times 2 \times 3 \times 4} \times \frac{8 \times 7 \times 6 \times 5}{1 \times 2 \times 3 \times 4}\)

= 35 × 70 = 2450

3. A bag contains 2 red, 3 green and 2 blue balls. 2 balls are to be drawn randomly. What is the probability that the balls drawn contain no blue ball?
(a) 57
(b) 1021
(c) 27
(d) 1121
(e) None of these
Ans.b

Reqd probability = \(\frac{5_{C_{2}}}{7_{C_{2}}} = \frac{5 \times 4}{7 \times 6} = \frac{10}{21}\)

4. In how many different ways can the letters of the word BOOKLET be arranged such that B and T always come together?
(a) 360
(b) 720
(c) 480
(d) 5040
(e) None of these
Ans.b

Treat B and T as a single letter. Then the remaining letters (5 + 1 = 6) can be arranged in 6! ways. Since, O is repeated twice, we have to divide by 2 and the B and T letters can be arranged in 2! ways.

Total no. of ways = \(\frac{6! \times 2!}{2} = 720\)

5. In a box there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?
(a) 719
(b) 23
(c) 34
(d) 921
(e) None of these
Ans.e

If the drawn ball is neither red nor green, then it must be blue, which can be picked in 7C1 = 7 ways. One ball can be picked from the total(8 + 7 + 6 = 21) in 21C1 = 21 ways.

∴ Reqd probability = 721 = 13

6. In how many different ways can the letters of the word RUMOUR be arranged?
(a) 180
(b) 720
(c) 30
(d) 90
(e) None of these
Ans.a

Reqd. number of ways

\(\frac{6!}{2! \times 2!} = \frac{6 \times 5 \times 4 \times 3}{1 \times 2} = 180\)
7. 765 chairs are to be arranged in a column in such a way that the number of chairs in each column should be equal to the columns. How many chairs will be excluded to make this arrangement possible?
(a) 6
(b) 36
(c) 19
(d) 27
(e) None of these
Ans.b

272 < 765 < 282

∴ required no. of chairs to be excluded

= 765 – 729 = 36

8. In how many different ways can the letters of the word JUDGE be arranged so that the vowels always come together?
(a) 48
(b) 24
(c) 120
(d) 60
(e) None of these
Ans.a

Reqd. number = 4! × 2! = 24 × 2 = 48

9. How many words can be formed from the letters of the word SIGNATURE so that the vowels always come together?
(a) 720
(b) 1440
(c) 3600
(d) 2880
(e) None of these
Ans.e

The word SIGNATURE consists of nine letters comprising four vowels (A, E, I and U) and five consonants (G, N, R, T and S). When the four vowels are considered as one letter, we have six letters which can be arranged in 6P6 ways i.e 6! ways. Note that the four vowels can be arranged in 4! ways.

Hence required number of words

= 6! × 4! = 720 × 24

= 17280

10. In how many ways a committee consisting of 5 men and 6 women can be formed from 8 men and 10 women?
(a) 266
(b) 86400
(c) 11760
(d) 5040
(e) None of these
Ans.c

Here, 5 men out of 8 men and 6 women out of 10 women can be chosen in

8C5 × 10C6 ways, i.e., 11760 ways.