# There are 10 persons named

Question:

There are 10 persons named $P_{1}, P_{2}, P_{3}, \ldots, P_{10}$. Out of 10 persons, 5 persons are to be arranged in a line such that in each arrangement $P_{1}$ must occur whereas $P_{4}$ and $P_{5}$ do not occur. Find the number of such possible arrangements.

Solution:

We need to arrange 5 persons in a line out of 10 persons, such that in each arrangement P1 must occur whereas P4 and P5 do not occur.

First we choose 5 persons out of 10 persons, such that in each arrangement P1 must occur whereas P4 and P5 do not occur.

Number of such selections = 7C4

Now, in each selection 5 persons can be arranged among themselves in 5! ways.

$\therefore$ required number of arrangements $={ }^{7} C_{4} \times 5 !=\frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times 5 \times 4 \times 3 \times 2 \times 1=4200$

Thus, ​number of such possible arrangements is 4200.