# Exercise : 3

(b) 72

(c) 288

(d) 720

(e) None of these

**Ans.d**

The first and the last (terminal) digits are even and there are three even digits. This arrangement can be done in ^{3}P_{2} ways. For any one of these arrangements,two even digits are used; and the remaining digits are 5 (4 odd and 1 even) and the four digits in the six digits(leaving out the terminal digits) may be arranged using these 5 digits in ^{5}P_{4} ways. The required number of numbers is ^{3}P_{2} × ^{5}P_{4} = 6 × 120 = 720.

(b) 36

(c) 125

(d) 91

(e) None of these

**Ans.d**

Required number of possible outcomes

= Total number of possible outcomes – Number of possible outcomes in which 5 does not appear on any dice. (hence 5 possibilities in each throw)

= 6^{3} – 5^{3} = 216 – 125 = 91

_{1}, A

_{2}, ...., A

_{10}can be ranked so that A

_{1}is always above A

_{2}is

^{10!}⁄

_{2}

(b) 10!

(c) 9!

(d)

^{8!}⁄

_{2}

(e) None of these

**Ans.a**

Ten candidates can be ranked in 10! ways. In half of these ways A_{1} is above A_{2} and in another half A_{2} is above A_{1}.

So, required number of ways is ^{10!}⁄_{2}

^{2}

(b) 2

^{n}

(c) 2n

(d) 3

^{n}

(e) None of these

**Ans.b**

Let the two boxes be B_{1} and B_{2}. There are two choices for each of the n objects. So, the total number of ways is

2 × 2 ×.....× 2 (n times ) = 2^{n}

(b) 576

(c) 440

(d) 144

(e) None of these

**Ans.b**

Required number of possible outcomes

= Total number of possible outcomes – Number of possible outcomes in which all vowels are together

= 6! – 4 ! × 3! = 576

(b) 180

(c) 192

(d) 360

(e) None of these

**Ans.a**

Task 1 can not be assigned to either person 1 or 2 i.e.there are 4 options.

Task 2 can be assigned to 3 or 4

So, there are only 2 options for task 2.

So required no. of ways = 2 options for task 2 × 3 options for task 1 × 4 options for task 3 × 3 options for task 4 × 2 options for task 5 × 1 option for task 6.

= 2 × 3 × 4 × 3 × 2 × 1 = 144

(b) 881

(c) 891

(d) 879

(e) None of these

**Ans.d**

The required number of ways

= (10 + 1)(9 + 1)(7 + 1) - 1 = 879.

(b) 100

(c) 120

(d) 240

(e) None of these

**Ans.c**

The number of ways of selecting 3 persons from 12 people under the given condition :

Number of ways of arranging 3 people among 9 people seated in a row, so that no two of them are consecutive

= Number of ways of choosing 3 places out of the 10[8 in between and 2 extremes]

= ^{10}C_{3} = \(\frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 5 \times 3 \times 8 = 120\)

^{10}

(b) 2

^{10}– 1

(c) 3

^{10}– 1

(d) 2

^{10}

(e) None of these

**Ans.c**

Since each question can be selected in 3 ways, by selecting it or by selecting its alternative or by rejecting it. Thus, the total number of ways of dealing with 10 given questions is 3^{10} including a way in which we reject all the questions.

Hence, the number of all possible selections is 3^{10} – 1.

(b) 126

(c) 91

(d) 94

(e) None of these

**Ans.c**

Number of ways of selecting 5 guests from nine friends = ^{9}C_{5}

Out of these, ^{7}C_{3} ways are those in which two of the friends occur together [3 more persons to be selected out of remaining 7]

∴ Number of ways, in which two of the friends will not attend the party together = ^{9}C_{5} – ^{7}C_{3} = 91.