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A lady gives a dinner party for six guests.the number of ways in which they may be selected from among ten friends,if two of the friends will not attend

the party together is



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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.

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