# Exercise : 1

(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.

(b) 105

(c) 1170

(d) Cannot be determined

(e) None of these

**Ans.a**

Reqd no. of ways = ^{7}C_{4} × ^{8}C_{4}

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

= 35 × 70 = 2450

^{5}⁄

_{7}

(b)

^{10}⁄

_{21}

(c)

^{2}⁄

_{7}

(d)

^{11}⁄

_{21}

(e) None of these

**Ans.b**

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

(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 = \(\frac{6! \times 2!}{2} = 720\)

^{7}⁄

_{19}

(b)

^{2}⁄

_{3}

(c)

^{3}⁄

_{4}

(d)

^{9}⁄

_{21}

(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 ^{7}C_{1} = 7 ways. One ball can be picked from the total(8 + 7 + 6 = 21) in ^{21}C_{1} = 21 ways.

∴ Reqd probability = ^{7}⁄_{21} = ^{1}⁄_{3}

(b) 720

(c) 30

(d) 90

(e) None of these

**Ans.a**

Reqd. number of ways

\(\frac{6!}{2! \times 2!} = \frac{6 \times 5 \times 4 \times 3}{1 \times 2} = 180\)(b) 36

(c) 19

(d) 27

(e) None of these

**Ans.b**

27^{2} < 765 < 28^{2}

∴ required no. of chairs to be excluded

= 765 – 729 = 36

(b) 24

(c) 120

(d) 60

(e) None of these

**Ans.a**

Reqd. number = 4! × 2! = 24 × 2 = 48

(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 ^{6}P_{6} 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

(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

^{8}C_{5} × ^{10}C_{6} ways, i.e., 11760 ways.