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  • If the letters of the word ASSASSINATION are arranged at random. Find the Probability that (a) Four S’s come consecutively in the word (b) Two I’s and two N’s come together (c) All A’s are not coming together (d) No two A’s are coming together.

If the letters of the word ASSASSINATION are arranged at random. Find the Probability that
(a) Four S’s come consecutively in the word
(b) Two I’s and two N’s come together
(c) All A’s are not coming together
(d) No two A’s are coming together.

Answers (1)

Given word: ASSASINATION

Total no. of letters in the word = 13

Viz., 3 A’s, 4 S’s, 2 I’s, 1 T & 1 O

No. of ways in which these letters can be arranged –

n(S) = 13! / 3! 4! 2! 2!

  1. 4 S’s come consecutively in ASSASINATION

Then the word becomes-

S S S S A A I N A T I O N

No. of letters = 1 + 9 = 10Now,

Thus,

n(E) = 10! / 3! 2! 2!

Now,

Required probability = \frac{\frac{10!}{3!2!2!}}{\frac{13!}{3!4!2!2!}}

= 10! / 3! 2! 2!  x  3! 4! 2! 2! / 13!

= 10! x 4!/ 13 x 12 x 11 x 10!

= 2 /143

  1. 2 I’s & 2 N’s come together

Then the word becomes-

I I N N A S S A S S A T O


No. of letters = 1 + 9 = 10Now,

Thus,

n(E) = 4! / 2! 2!  x  10!/ 3! 4!

Now,

Required probability = \frac{\frac{4!10!}{3!4!2!2!}}{\frac{13!}{3!4!2!2!}}

= 10! 4! / 3! 4! 2! 2!  x  3! 4! 2! 2! / 13!

= 10! x 4!/ 13 x 12 x 11 x 10!

= 2 /143

  1. All A’s are not coming together

Then the word becomes-

A A A S S S S I N T I O N

Now,
No. of letters = 1 + 10 = 11         

Thus,

When all A’s come together no. of words = 11! / 4! 2! 2!

Now,

probability = \frac{\frac{11!}{4!2!2!}}{\frac{13!}{3!4!2!2!}}

= 11! / 4! 2! 2!  x  3! 4! 2! 2! / 13!

= 11! x 3!/ 13 x 12 x 11!

= 1 / 26

Now,

P (All A’s don’t come together) = 1 – P (all A’s comes together)

                              = 1 – 1/26

                              = 25 / 26

  1. No two A’s are coming together,

Then the word becomes-

S   S   S   S   I   N   T   I   O   N


No. of ways of arranging except A = 10! / 4! 2! 2!Now,

There are 11 vacant places

Total no. of A’s in ASSASINATION = 3

Thus, the 3 A’s can be placed in 11C3 ways

= 11! / 3! (11 – 3)!

= 11! / 3! 8!

No. of ways when 2 A’s are not together

= 11! / 3! 8!  x  10! / 4! 2! 2!  x  3! 4! 2! 2!/ 13!

= 11! X 10 x 9 x 8! / 8! x 13 x 12 x 11!

= 10 x 9 / 13 x 12

= 15 / 26

Posted by

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