Question

There are 10 persons named $P_1,P_2,P_3,....,P_{10}$. Out of 10 persons, 5 persons are to be arranged in a line such that in each arrangement P₁ must occur whereas P₄ and P₅ do not occur. Find the number of such possible arrangements.

Answer 1

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.

∴ required number of arrangements = ⁷C₄ × 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.