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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 suchthat is each arrangement P1 must occur whereas P4 andP5 do not occur. Find the number of such possible arrangements.



Answer -

Given:

Total persons = 10

Number of persons tobe selected = 5 from 10 persons (P1, P2, P3 …P10)

It is also told that P1 shouldbe present and P4 and P5 should not be present.

We have to choose 4persons from remaining 7 persons as P1 is selected and P4 andP5 are already removed.

Number of ways =Selecting 4 persons from remaining 7 persons

7C4

By using the formula,

nCr = n!/r!(n – r)!

7C4 = 7! / 4!(7 – 4)!

= 7! / (4! 3!)

= [7×6×5×4!] / (4! 3!)

= [7×6×5] / (3×2×1)

= 7×5

= 35

Now we need to arrangethe chosen 5 people. Since 1 person differs from other.

35 × 5! = 35 ×(5×4×3×2×1)

= 4200

 The total no. ofpossible arrangement can be done is 4200.

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