Question

In a circus, there are ten cages for accommodating ten animals. Out of these four cages are so small that five out of $10$ animals cannot enter them. In how many ways will it be possible to accommodate ten animals in these ten cages?

• A. $66400$
• B. $86400$
• C. $96400$
• D. None of these

Answer 1

The correct option is B

$86400$

Explanation for correct option:

Finding the number of ways:

We have to first accommodate these $5$ animals in cages that can't enter in $4$ small cages,

Number of possible ways to do so =${}^{6}P_5$

Number of possible ways to accommodate $5$ animals in $5$ cages = $5!$

Now, total number of ways possible,

$$\begin{gathered} {}^{6}P_{5}x5!=6!X5! \\ =86400 \end{gathered}$$

Hence, the correct option is B.