Question

A man has three friends. The number of ways he can invite one friend every day for dinner on six successive nights so that no friend is invited more than three times is
(a) 640
(b) 320
(c) 420
(d) 510

Answer 1

Hint: The possible situations are 3 for which the conditions satisfy. For the 1st given situation one often will be invited for three days, in between remaining 2 one of them will be invited for 2 days and the other for remaining 1 day. Now, for the 2nd situation all are invited for 2 days each. And in the last situation any two of the three will be invited for 3 days each. Then find the ways of all the situations hence, add them all.

Complete step-by-step answer:
In the question, we are told that a man has three friends. Now we have to find the number of ways to invite one friend everyday for dinner on six successive nights so that no friend is invited more than three times.
After observing the question and analyzing the situation we get that there are three different possible situations where three possible can be invited such that no one is invited more than three times.
So, let’s go for 1^st situation, one of the three will be invited once, in between rest of two anyone will be invited twice and last will be invited thrice.
Anybody among three can be invited in 3! or 6 ways.
So, the person coming for three days can be chosen in \[{}^{6}{{C}_{3}}\] ways. So, there are 3.
The person coming for two days can be chosen in \[{}^{3}{{C}_{2}}\] ways. And the last one will come on the left 1 day.
Hence, the total number of ways is \[{}^{6}{{C}_{3}}\times {}^{3}{{C}_{2}}\times {}^{1}{{C}_{1}}\times 6\] . Now as we know that ${}^{n}{{C}_{r}}$ is equal to $\dfrac{n!}{\left( n-r \right)!\times r!}$ so we can calculate it as,
\[\dfrac{6!}{\left( 6-3 \right)!\times 3!}\times \dfrac{3!}{\left( 3-2 \right)!\times 2!}\times \dfrac{1!}{0!\times 1!}\times 6\]
\[\Rightarrow \dfrac{720}{6\times 6}\times \dfrac{6}{1\times 2}\times \dfrac{1}{1}\times 6\]
\[\Rightarrow 360\] ways
Now let’s go to the 2^nd situation where all the friends were invited two days each.
All are invited in 1 way.
So, the person coming for two days can be chosen in \[{}^{6}{{C}_{2}}\] ways. So, there are 4. The person coming for two days can be chosen in \[{}^{4}{{C}_{2}}\] ways. And the last one will come on the left 2 days.
As the distribution is the same, we will divide by 3! As total persons are three, we will multiply it by 3.
Hence the total number of ways is $\dfrac{\left( {}^{6}{{C}_{2}}\times {}^{4}{{C}_{2}}\times {}^{2}{{C}_{1}}\times 3 \right)}{3!}$ . Now as we know that ${}^{n}{{C}_{r}}$ is equal to $\dfrac{n!}{\left( n-r \right)!\times r!}$ so we can calculate it as,
\[\dfrac{1}{3!}\times \dfrac{6!}{\left( 6-2 \right)!\times 2!}\times \dfrac{4!}{\left( 4-2 \right)!\times 2!}\times \dfrac{2!}{\left( 2-1 \right)!\times 1!}\times 3\]
\[\Rightarrow \dfrac{720}{24\times 2}\times \dfrac{24}{2\times 2}\times \dfrac{2}{1}\times 3\times \dfrac{1}{6}\]
\[\Rightarrow 90\] ways
Now, let’s go to the last situation where only 2 of the 3 were invited on three days each.
So now choosing 2 friends from 3 friends, it can be done in ${}^{3}{{C}_{2}}$ or 3 ways.
So, the person coming for three days can be chosen in ${}^{6}{{C}_{3}}$ ways. So, now left days are 3.
And the last one will come on the left days in ${}^{3}{{C}_{3}}$ ways.
Hence, the total number of ways is $\left( {}^{6}{{C}_{3}}\times {}^{3}{{C}_{3}}\times 3 \right)$ . Now as we know that ${}^{n}{{C}_{r}}$ is equal to $\dfrac{n!}{\left( n-r \right)!\times r!}$ so we can calculate it as,
$\dfrac{6!}{\left( 6-3 \right)!\times 3!}\times \dfrac{3!}{\left( 3-3 \right)!\times 3!}\times 3$
$\Rightarrow \dfrac{720}{6\times 6}\times \dfrac{6}{1\times 6}\times 3$
$\Rightarrow 60$ ways
So, the total number of ways is 360+90+60 ways or 510 ways.
So, the correct option is ‘d’.

Note: Students can also do the question by finding all possible ways and then subtracting those cases which does not satisfy the situation given in the question.