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# Six boys and six girls sit in a row at random. Find the probability th

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Six boys and six girls sit in a row at random. Find the probability th  [#permalink]

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28 Jan 2019, 04:47
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Difficulty:

55% (hard)

Question Stats:

40% (01:59) correct 60% (02:09) wrong based on 26 sessions

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Six boys and six girls sit in a row at random. Find the probability that the boys and girls sit alternately.

A. 1/66

B. 1/132

C. 1/231

D. 1/462

E. 1/623

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Re: Six boys and six girls sit in a row at random. Find the probability th  [#permalink]

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28 Jan 2019, 05:46
Bunuel wrote:
Six boys and six girls sit in a row at random.Find the probability that the boys and girls sit alternately.

A. 1/66

B. 1/132

C. 1/231

D. 1/462

E. 1/623

If the boys and girls sit alternatively
The required probability in that case would be given by
= 2 * 6! * 6!/ 12!
= 1/462.
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Re: Six boys and six girls sit in a row at random. Find the probability th  [#permalink]

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Updated on: 29 Jan 2019, 00:25
Total no. of ways of arranging 12 people are 12! ways

(i) _B_B_B_B_B_B_

6 boys can be arranged in 6! ways. Now, these 6 girls can be arranged in 6! ways.

Total possible arrangements are $$6!*6!$$
There are such cases possible in this manner.

So total possible cases are $$2*6!*6!$$

Required probability is $$\frac{2*6!*6!}{12!}$$

$$\frac{2*2*3*4*5*6}{7*8*9*10*11*12}$$

$$\frac{3*4}{7*8*9*11}$$

$$\frac{1}{42*11}$$

$$\frac{1}{462}$$

OPTION: D
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Originally posted by eswarchethu135 on 28 Jan 2019, 07:13.
Last edited by eswarchethu135 on 29 Jan 2019, 00:25, edited 1 time in total.
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Re: Six boys and six girls sit in a row at random. Find the probability th  [#permalink]

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28 Jan 2019, 07:55
Total 6 boys and 6 girls :12!
Boy and girl alternative: 6! *6! *2 ways
boy first 6! *6! or girl first 6! *6!.
= 2 . 6! . 6!/ 12!
= 1/462
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Re: Six boys and six girls sit in a row at random. Find the probability th  [#permalink]

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30 Jan 2019, 18:12
Bunuel wrote:
Six boys and six girls sit in a row at random. Find the probability that the boys and girls sit alternately.

A. 1/66

B. 1/132

C. 1/231

D. 1/462

E. 1/623

We have the following two arrangements:

B - G - B - G - B - G - B - G - B - G - B - G

or

G - B - G - B - G - B - G - B - G - B - G - B

Each arrangement will yield the same result, so we have:

[(6 x 6 x 5 x 5 x 4 x 4 x 3 x 3 x 2 x 2 x 1 x 1) x 2. This is the number of ways that we can have alternating boys and girls in the seating arrangement.

We can simplify this as (6! x 6! x 2).

We now note that the TOTAL number of ways of seating the 12 children is 12!

Thus, the probability that the boys and girls sit alternately is:

(6! x 6! x 2)/12!

Simplifying, we have:

(6 x 5 x 4 x 3 x 2 x 2)/(12 x 11 x 10 x 9 x 8 x 7) =1/(11 x 3 x 2 x 7) = 1/462

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Re: Six boys and six girls sit in a row at random. Find the probability th   [#permalink] 30 Jan 2019, 18:12
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