If n is the number of ways five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then n is equal to:

Question

Aicharya · Accepted Answer

Given,

Rooms are identical,

So, total ways to distribute $5$ employees into $4$ rooms are:

$5, 0, 0, 0 \Rightarrow 1$ way

$4, 1, 0, 0 \Rightarrow \frac{5!}{4!} = 5$ ways

$3, 2, 0, 0 \Rightarrow \frac{5!}{3! 2!} = 10$ ways

$2, 2, 0, 1 \Rightarrow \frac{5!}{2! 2! 2!} = 15$ ways

$2, 1, 1, 1 \Rightarrow \frac{5!}{2! {(1!)}^{3} 3!} = 10$ ways

$3, 1, 1, 0 \Rightarrow \frac{5!}{3! 2!} = 10$ ways

Total $\Rightarrow 1 + 5 + 10 + 15 + 10 + 10 = 51$ ways

If n is the number of ways five different employees can sit into four indistinguishable offices where any…

Solution