A lady giving a dinner party 46 guest the number of ways of selecting from 10 friends in which two friends not standing together
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Let us say that the two particular friends are A and B.
If A is invited among six guests and B is not, then: number of combinations to select 5 more guests from the remaining 8 friends:
C(8, 5) = 8 ! / (5! 3!) = 56
If B is invited among the six guests and A is not , then the number of ways of selecting the remaining 5 guests = C(8, 5) = 56
Suppose both A and B are not included in the six guests list : then the number of such combinations = C(8, 6) = 7 * 8 /2 = 28
So the total number of sets of guests that can be selected = 140.
If A is invited among six guests and B is not, then: number of combinations to select 5 more guests from the remaining 8 friends:
C(8, 5) = 8 ! / (5! 3!) = 56
If B is invited among the six guests and A is not , then the number of ways of selecting the remaining 5 guests = C(8, 5) = 56
Suppose both A and B are not included in the six guests list : then the number of such combinations = C(8, 6) = 7 * 8 /2 = 28
So the total number of sets of guests that can be selected = 140.
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total no of guests are 140
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