Question

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
A. 112
B. 140
C. 164
D. none of these

Answer 1

We first separate the events in which the people can be in the party keeping up with the given condition. We separately find the number of ways we can arrange the people and add them to find the final number of possible ways.

Complete step by step answer:
A lady gives a dinner party for six guests out of 10 people. The condition being 2 out of those 10 will not attend the party together. So, we can get two situations. One being any one of them attends the party. The second one is that none of them attend the party.
We separately find the number of ways. We take the first situation where only one attends.
Here, we can choose one of them in ${}^{2}{{C}_{1}}=2$ ways. The choice was for 6 people out of 10 people and now it became 5 out of $10-2=8$ people.
The number of choices will be ${}^{8}{{C}_{5}}=\dfrac{8!}{5!\times 3!}=56$ ways.
So, the total number of ways will be $56\times 2=112$.
Now we take the second situation where none attends.
Here, the choice was for 6 people out of 10 people and now it became 6 out of $10-2=8$ people.
The number of choices will be ${}^{8}{{C}_{6}}=\dfrac{8!}{6!\times 2!}=28$ ways.
Total will be $112+28=140$.

So, the correct answer is “Option B”.

Note: We also could have separated the first event in person based but the number of ways is same for both of them and that’s why we multiplied 2 with the value to get the total. There are some constraints in the form of ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}$. The general conditions are $n\ge r\ge 0;n\ne 0$.