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# From 8 people, a team of 4 needs to be selected. How many ways are pos

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Re: From 8 people, a team of 4 needs to be selected. How many ways are pos [#permalink]
I solved by using the method in which we have:-

A and B not together --> (Total possibilities)-(When A and B are selected together)

Total Possibilities:-8C4
Case when A and B both are selected:-
In this case 2 places are occupied by A and B. Rest 2 can be filled in 6C4 ways.

=70-15
=55

Therefor B is correct.
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Re: From 8 people, a team of 4 needs to be selected. How many ways are pos [#permalink]
Bunuel wrote:
SPHOORTHI wrote:
From 8 people, a team of 4 needs to be selected. How many ways are possible if two particular people do not want to be together in the team?

(A) 35
(B) 55
(C) 40
(D) 60
(E) 75

Say A and B do not want to be together in the team.

The number of teams with A but not B = 6C3 = 20 (we have A in the team and choosing the remaining 3 members out of 8 - A - B = 6)

The number of teams with B but not A = 6C3 = 20 (we have B in the team and choosing the remaining 3 members out of 8 - A - B = 6)

The number of teams without both A and B = 6C4 = 15 (choosing all 4 members from 8 - A - B = 6).

Total = 20 + 20 + 15 = 55.

But how do we know that it is A and B, and not C, D E F etc.
Don’t we need to select them too?

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Re: From 8 people, a team of 4 needs to be selected. How many ways are pos [#permalink]
Stanindaw

No, we are supposed to keep two particular people from being together, and those can be represented as A and B. Surely, in a team of 4, there will be plenty of pairs of people who ARE together. If we had to exclude them all, the answer would be 0.
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Re: From 8 people, a team of 4 needs to be selected. How many ways are pos [#permalink]
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Re: From 8 people, a team of 4 needs to be selected. How many ways are pos [#permalink]
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