In the below word how many words are there in whic | Permutations and Combinations Questions & Answers

Q:

In how many ways the letters of the word 'DESIGN' can be arranged so that no consonant appears at either of the two ends?

A) 240	B) 72
C) 48	D) 36

Answer & Explanation Answer: C) 48

Explanation:

DESIGN = 6 letters

No consonants appear at either of the two ends.

= $2 x 4 P_{4}$ = 2 x 4 x 3 x 2 x 1= 48

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10 9566

Q:

In How many ways can the letters of the word 'CAPITAL' be arranged in such a way that all the vowels always come together?

A) 360	B) 720
C) 120	D) 840

Answer & Explanation Answer: A) 360

Explanation:

CAPITAL = 7

Vowels = 3 (A, I, A)

Consonants = (C, P, T, L)

5 letters which can be arranged in $5 P_{5} = 5!$

Vowels A,I = $\frac{3!}{2!}$

No.of arrangements = 5! x $\frac{3!}{2!}$ =360

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4 5442

Q:

The number of ways that 8 beads of different colours be strung as a necklace is

A) 2520	B) 2880
C) 4320	D) 5040

Answer & Explanation Answer: A) 2520

Explanation:

The number of ways of arranging n beads in a necklace is $\frac{(n - 1)!}{2} = \frac{(8 - 1)!}{2} = \frac{7!}{2}$ = 2520

(since n = 8)

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27 16096

Q:

The number of ways that 7 teachers and 6 students can sit around a table so that no two students are together is

A) 7! x 7!	B) 7! x 6!
C) 6! x 6!	D) 7! x 5!

Answer & Explanation Answer: B) 7! x 6!

Explanation:

The students should sit in between two teachers. There are 7 gaps in between teachers when they sit in a roundtable. This can be done in $7 P_{6}$ ways. 7 teachers can sit in (7-1)! ways.

Required no.of ways is = $7 P_{6} . 6! = 7! . 6!$

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10 5564

Q:

If A1, A2, A3, A4, ..... A10 are speakers for a meeting and A1 always speaks after, A2 then the number of ways they can speak in the meeting is

A) 9!	B) 9!/2
C) 10!	D) 10!/2

Answer & Explanation Answer: D) 10!/2

Explanation:

As A1 speaks always after A2, they can speak only in 1st to 9th places and

A2 can speak in 2nd to 10 the places only when A1 speaks in 1st place

A2 can speak in 9 places the remaining

A3, A4, A5,...A10 has no restriction. So, they can speak in 9.8! ways. i.e

when A2 speaks in the first place, the number of ways they can speak is 9.8!.

When A2 speaks in second place, the number of ways they can speak is 8.8!.

When A2 speaks in third place, the number of ways they can speak is 7.8!. When A2 speaks in the ninth place, the number of ways they can speak is 1.8!

Therefore,Total Number of ways they can speak = (9+8+7+6+5+4+3+2+1) 8! = $\frac{9}{2} (9 + 1) 8!$ = 10!/2

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5 4230

Q:

In a plane 8 points are colliner out of 12 points, then the number of triangles we get with those 12 points is

A) 20	B) 160
C) 164	D) 220

Answer & Explanation Answer: C) 164

Explanation:

For a triangle, we need 3 non-collinear points. So with 12 points (when all the 12 are such that any three non-collinear is $12 C_{3}$ . But among them 8 points are collinear.

If all these 8 points are different we get $8 C_{3}$ triangles as they are collinear.

In $12 C_{3}$ triangles, we do not get $8 C_{3}$ triangles

Therefore, The number of triangles we get = $12 C_{3} - 8 C_{3}$ = 164

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3 3658

Q:

Find the value of 'n' for which the nth term of two AP'S:

15,12,9.... and -15,-13,-11...... are equal?

A) n = 2	B) n = 5
C) n = 29/5	D) n = 1

Answer & Explanation Answer: C) n = 29/5

Explanation:

Given are the two AP'S:

15,12,9.... in which a=15, d=-3.............(1)

-15,-13,-11..... in which a'=-15 ,d'=2.....(2)

now using the nth term's formula,we get

a+(n-1)d = a'+(n-1)d'

substituting the value obtained in eq. 1 and 2,

15+(n-1) x (-3) = -15+(n-1) x 2

=> 15 - 3n + 3 = -15 + 2n - 2

=> 12 - 3n = -17 + 2n

=> 12+17 = 2n+3n

=> 29=5n

=> n= 29/5

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19 51534

Q:

Find the number of ways in which 21 balls can be distributed among 3 persons such that each person does not receive less than 5 balls.

A) 28	B) 14
C) 21	D) 7

Answer & Explanation Answer: A) 28

Explanation:

Let x, y, z be the number of balls received by the three persons, then

$x \geq 5, y \geq 5, z \geq 5 a n d x + y + z = 21$

Let $u \geq 0, v \geq 0, w \geq 0 t h e n$

$∴$ x + y + z =21

$\Rightarrow$ u + 5 + v + 5 + w + 5 = 21

$\Rightarrow$ u + v + w = 6

$∴$ Total number of solutions = ${}^{6 + 3 - 1}C_{3 - 1} = {}^{8}C_{2} = 28$

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18 7972

A) 120	B) 180
C) 210	D) 240

In the below word how many words are there in which R and W are at the end positions?

RAINBOW

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In the below word how many words are there in which R and W are at the end positions? RAINBOW

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In the below word how many words are there in which R and W are at the end positions?

RAINBOW