There are 10 persons named P1, P2, P3 …, P10. Out of 10 persons, 5 persons are to be arranged in a line such that is each arrangement P1 must occur whereas P4 and P5 do not occur. Find the number of such possible arrangements.

Asked by Aaryan | 1 year ago |  225

##### Solution :-

Given:

Total persons = 10

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

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

We have to choose 4 persons from remaining 7 persons as P1 is selected and P4 and P5 are already removed.

Number of ways = Selecting 4 persons from remaining 7 persons

7C4

By using the formula,

nCr =$$\dfrac{ n!}{r!(n – r)!}$$

7C4 =$$\dfrac{ 7! }{ 4!(7 – 4)!}$$

= $$\dfrac{ 7! }{ (4! 3!)}$$

= 7×5

= 35

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

35 × 5! = 35 × (5×4×3×2×1)

= 4200

The total no. of possible arrangement can be done is 4200.

Answered by Sakshi | 1 year ago

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