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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.
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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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