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Probability question

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Probability question  [#permalink]

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New post 06 Mar 2018, 23:25
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Hello guys,

sometimes I struggle with that type of question.

Q : If among 5 children there are 2 siblings, in how many ways can the children be seated in a row so that the siblings do not sit together?

A : 72. Without limitations, 5 children can be seated in 5!=120 ways. Find the number of ways to seat the 5 children so that the siblings DO sit together. The siblings can be regarded as one unit so there are 4! combinations. But within this unit the siblings can sit in two different ways. So the number of ways to seat the 5 children so that the siblings DO sit together is 4!∗2=48. Thus, the number of combinations in which the siblings DO NOT sit together is 120 - 48 = 72.

So If the question was If among 5 children there are 3 siblings, the answer will be 120-(4!∗3), am I right?

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Re: Probability question  [#permalink]

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New post 06 Mar 2018, 23:43
thegame12 wrote:
Hello guys,

sometimes I struggle with that type of question.

Q : If among 5 children there are 2 siblings, in how many ways can the children be seated in a row so that the siblings do not sit together?

A : 72. Without limitations, 5 children can be seated in 5!=120 ways. Find the number of ways to seat the 5 children so that the siblings DO sit together. The siblings can be regarded as one unit so there are 4! combinations. But within this unit the siblings can sit in two different ways. So the number of ways to seat the 5 children so that the siblings DO sit together is 4!∗2=48. Thus, the number of combinations in which the siblings DO NOT sit together is 120 - 48 = 72.

So If the question was If among 5 children there are 3 siblings, the answer will be 120-(4!∗3), am I right?


Discussed here: m08-183801.html

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Re: Probability question &nbs [#permalink] 06 Mar 2018, 23:43
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