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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.
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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.
∴ 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.
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