Question

A man has three friends. In how many ways he can invite one friend everyday for dinner on 6 successive nights so that no friend is invited more than 3 times.
(a) 600
(b) 510
(c) 255
(d) 592

Answer 1

Hint: First, we assume the three events are a, b and c times in which x, y and z are invited. Then, we have the following possibilities for the cases that no friend is selected more than 3 times as (1,2,3) or (3,3,0) or (2,2,2). Then, by using them, we get an expression as $\dfrac{6!}{1!2!3!}\times 3!+\dfrac{6!}{3!3!2!}\times 3!+\dfrac{6!}{2!2!2!}\times 3!$.Then, by using the concept of factorial, we get the desired result.

Complete step-by-step answer:
In this question, we are supposed to find out how many ways a person can invite one friend everyday for dinner on 6 successive nights so that no friend is invited more than 3 times from 3 friends he has.
Now, let us assume the three friends are x, y and z.
Then, we assume the three events are a, b and c times in which x, y and z are invited.
So, it denotes a case in which x is invited a time.
Similarly, b denotes a case in which y is invited b times.
Also, c denotes a case in which z is invited z times.
Now, we have following possibilities for the cases that no friend is selected more than 3 times as:
(a, b, c)=(1,2,3) or (3,3,0) or (2,2,2)
This shows the grouping of the invitees of 6 days per week.
Now, by using the factorial ways to get the above condition being successful, we need to understand how to calculate factorials.
To find the factorial of the number n, multiply the number n with (n-1) till it reaches 1. To understand let us find the factorial of 4.
$\begin{align}
  & 4!=4\times 3\times 2\times 1 \\
 & \Rightarrow 24 \\
\end{align}$
So, the total number of the ways is given by the expression as:
$\dfrac{6!}{1!2!3!}\times 3!+\dfrac{6!}{3!3!2!}\times 3!+\dfrac{6!}{2!2!2!}\times 3!$
Now, we will solve the above expression to get the total number of ways as:
$\begin{align}
  & \dfrac{6\times 5\times 4\times 3!}{1!2!3!}\times 3!+\dfrac{6\times 5\times 4\times 3!}{3!3!2!}\times 3!+\dfrac{6\times 5\times 4\times 3\times 2!}{2!2!2!}\times 3! \\
 & \Rightarrow 360+60+90 \\
 & \Rightarrow 510 \\
\end{align}$
So, 510 total ways are there for a person inviting one friend for dinner on 6 successive nights without having a friend more than 3 times.
Hence, option (b) is correct.

Note: In this type of questions, we will mainly use the concept of the factorial to get the number of ways by using selection. Moreover, we should be good at assuming things as in this question also we have assumed the three friends as x, y ,z and even assumed their frequency of being invited as a, b ,c respectively. So, the concept of the factorial is as follows:
To find the factorial of the number n, multiply the number n with (n-1) till it reaches 1. To understand let us find the factorial of 4.
$\begin{align}
  & 4!=4\times 3\times 2\times 1 \\
 & \Rightarrow 24 \\
\end{align}$