A lady gives a dinner party to 5 guests to be selected from 9 friends. The number of ways of forming the party of 5 given that two of the friends will not attend the party together is

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126

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None of these

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Solution

The correct option is C 91
There are $9$ friends

If two of term do not attend the party together

Total no. of ways is select $5$ from $=^{7} C_{5}$
$= 21$ ways

If one of them is selected for party, then no. of ways $=^{2} C_{1} \times^{7} C_{4} = 70$ ways

Total no. of ways $= 21 + 70 = 91$ ways.

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A lady gives a dinner party to 5 guests to be selected from 9 friends. The number of ways of forming the party of 5 given that two of the friends will not attend the party together is

The correct option is C 91There are 9 friendsIf two of term do not attend the party togetherTotal no. of ways is select 5 from =7C5=21 waysIf one of them is selected for party, then no. of ways=2C1×7C4=70 waysTotal no. of ways=21+70=91 ways.