# Exercise : 2

^{12}⁄

_{13}

(b)

^{11}⁄

_{13}

(c)

^{10}⁄

_{15}

(d)

^{1}⁄

_{15}

(e) None of these

**Ans.b**

Total possible ways of selecting 4 students out of 15 students = ^{15}C_{4} = \(\frac{15 \times 14 \times 13 \times 12}{1 \times 2 \times 3 \times 4} = 1365\)

The no. of ways of selecting 4 students in which no student belongs to Karnataka = ^{10}C_{4}

∴ Hence no. of ways of selecting at least one student from Karnataka = ^{15}C_{4} – ^{10}C_{4} = 1155

∴ Probability = ^{1155}⁄_{1365} = ^{77}⁄_{91} = ^{11}⁄_{13}

(b) 720

(c) 148

(d) 240

(e) None of these

**Ans.d**

Assume the 2 given students to be together (i.e one).

Now there are five students.

Possible ways of arranging them are = 5! = 120

Now, they (two girls) can arrange themselves in 2! ways.

Hence total ways = 120 × 2 = 240

(b) 60

(c) 48

(d) 32

(e) None of these

**Ans.e**

3 vowels can be arranged in three odd places in 3! ways.

Similarly, 3 consonants can be arranged in three even places in 3! ways.

Hence, the total number of words in which vowels occupy odd positions = 3! × 3! = 6 × 6 = 36 ways.

^{1}⁄

_{55}

(b)

^{54}⁄

_{55}

(c)

^{45}⁄

_{55}

(d)

^{3}⁄

_{55}

(e) None of these

**Ans.b**

n(S) = ^{12}C_{3} = \(\frac{12 \times 11 \times 10}{3 \times 2} = 2 \times 11 \times 10 = 220\)

No. of selection of 3 oranges out of the total 12 oranges

= ^{12}C_{3} = 2 × 11 × 10 = 220

No. of selection of 3 bad oranges out of the total 4 bad oranges = ^{4}C_{3} = 4

\n(E) = no. of desired selection of oranges

= 220 – 4 = 216

\ P(E) = \(\frac{n\left ( E \right )}{n\left ( S \right )} = \frac{216}{220} = \frac{54}{55}\)

(b) 2720

(c) 2160

(d) 1120

(e) None of these

**Ans.c**

Taking all vowels (IEO) as a single letter (since they come together) there are six letters

Hence no. of arrangements = ^{6!}⁄_{2!} × 3! = 2160

^{13}⁄

_{44}

(b)

^{41}⁄

_{44}

(c)

^{13}⁄

_{55}

(d)

^{52}⁄

_{55}

(e) None of these

**Ans.b**

Total no. of ways of drawing 3 marbles

= ^{12}C_{3} = \(\frac{12 \times 11 \times 10}{1 \times 2 \times 3} = 220\)

Total no. of ways of drawing marbles, which are of same colour = ^{5}C_{3} + ^{4}C_{3} + ^{3}C_{3} = 10 + 4 + 1 = 15

\Probability of same colour = ^{15}⁄_{220} = ^{3}⁄_{44}

\ Probability of not same colour = 1 - ^{3}⁄_{44} = ^{41}⁄_{42}

(b) 5040

(c) 1260

(d) 1200

(e) None of these

**Ans.c**

Possible arrangements are :

\(\frac{7!}{2! \times 2!} = 1260\) [division by 2 times 2! is because of the repetition of E and R](b) 162

(c) 168

(d) 163

(e) None of these

**Ans.b**

When 0 is the repeated digit like

100, 200, ...., 9 in number

When 0 occurs only once like

110, 220, ....., 9 in number

When 0 does not occur like

112,211, ....., 2 × (8 × 9) = 144 in number.

Hence, total = 9 + 9 + 144 = 162.

(b) 4500000

(c) 8100000

(d) 4400000

(e) None of these

**Ans.b**

Suppose x_{1} x_{2} x_{3} x_{4} x_{5} x_{6} x_{7} represents a seven digit number. Then x_{1} takes the value 1, 2, 3, ....., 9 and x_{2}, x_{3},....., x_{7} all take values 0, 1, 2, 3, ......, 9.

If we keep x_{1}, x_{2}, ......, x_{6} fixed, then the sum x_{1} + x_{2} + ......+ x_{6} is either even or odd. Since x_{7} takes 10 values 0, 1, 2, ....., 9, five of the numbers so formed will have sum of digits even and 5 have sum odd.

Hence the required number of numbers

= 9 . 10 . 10 . 10 . 10 . 10 . 5 = 4500000.

^{20}

(b) 5

^{20}– 1

(c) 5

^{20}+ 1

(d) 6

^{20}

(e) None of these

**Ans.a**

Number of single digit numbers = 5

Number of two digits numbers = 4 × 5

∵ [0 cannot occur at first place and repetition is allowed]

Number of three digits numbers

= 4 × 5 × 5 = 4 × 5^{2}

.... .... .... ....

.... .... .... ....

Number of 20 digits numbers = 4 × 5^{19}

∴ Total number of numbers

= 5 + 4.5 + 4.5^{2} + 4.5^{3}........4.5^{19}

= 5 + \(4 \cdot \frac{5\left ( 5^{19} - 1 \right )}{5 - 1}\) = 5 + 5^{20} - 5 = 5^{20}