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Question
There are 10 persons named P1, P2, P3, ... P10. 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.
Solution
Given that P1, P2, P3, ... P10 are 10 persons out of which 5 persons are to be arranged but P1 must occur and P4 and P5 never occur
∴ Selection is to be done only for 10 – 3 = 7 persons
∴ Number of selection = 7C4
= `(7!)/(4!(7 - 4)!)`
= `(7!)/(4!3!)`
= `(7*6*5*4!)/(4!*3*2*1)`
= 35
5 people can be arranged as 5!
So, the number of arrangement = 35 × 5!
= 35 × 120
= 4200
Hence, the required arrangement = 4200.
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