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.