Question

#Jee advanced question #

A man has 3 friends . if N is the number of ways he can invite everyday for dinner on 6 successive nights so that no friend is invited more than 3 times , then the value of N ​

Answer 1

Total no. of ways of inviting the 3 friends without any condition on 6 nights $=3^6=729.$

Condition is that no friend is invited more than 3 times.

So we've to deduct no. of ways of inviting friends among whom one is invited 4 or 5 or 6 times, from total no. of ways (729).

Case 1:- One among the 3 friends is invited 4 times.

Here we've to choose one among the 3 friends, to invite 4 times.

No. of ways of this combination $=\,^3\!C_1=3.$

Case 1.1:- Same person is chosen to invite on the other 2 nights.

One among the 3 friends is already invited 4 times, so we've to choose one among the other 2 friends to invite on the other 2 nights.

No. of ways of this $=\,^2\!C_1=2$

So total no. of such a combination $=3\times2=6.$

Combination is done, now we've to arrange them.

A friend is invited 4 times and another friend, 2 times. So the total no. of arrangements $=\dfrac{6!}{4!\cdot2!}=15.$

Hence total no. of ways $=6\times15=90.$

Case 1.2:- The two other friends are invited on the other 2 nights each.

No. of ways of choosing the other two friends $=\,^2\!C_2=1.$

So total no. of such a combination $=3\times1=3.$

Combination is done, now we've to arrange them.

Here, a friend is invited 4 times, and the other 2 are invited on other 2 nights each. So the total no. of arrangements $=\dfrac{6!}{4!}=30.$

Hence total no. of ways $=3\times30=90.$

Therefore, total no. of ways of inviting the friends among whom one is invited 4 times $=90+90=180.$

Case 2:- One among the 3 friends is invited 5 times.

No. of ways of choosing one among the 3 friends to invite 5 times $=\,^3\!C_1=3.$

And one among the other 2 friends have to be chosen to invite on the other night.

No. of ways of this combination $=\,^2\!C_1=2.$

Then, total no. of combination $=3\times2=6.$

Now we've a friend invited 5 times and another friend invited on the other night. So the total no. of arrangements $=\dfrac{6!}{5!}=6.$

Therefore, total no. of ways of inviting the friends among whom one is invited 5 times $=6\times6=36.$

Case 3:- One among the 3 friends is invited 6 times.

No. of ways of choosing one among the 3 to invite 6 times $=\,^3\!C_1=3.$

A friend is invited 6 times here. So total no. of arrangements $=\dfrac{6!}{6!}=1.$

Therefore, total no. of ways of inviting the friends among whom one is invited 6 times $=3\times1=3.$

Total no. of ways of inviting the 3 friends in 6 nights such that at least one among them should be invited at least 4 times $=180+36+3=219.$

Finally, no. of ways of inviting the 3 friends such that no friend is invited more than 3 times $=729-219=\mathbf{510}.$

Hence 510 is the answer.