Given:

Total persons = 10

Number of persons to be selected = 5 from 10 persons (P_{1}, P_{2}, P_{3} … P_{10})

It is also told that P_{1} should be present and P_{4} and P_{5} should not be present.

We have to choose 4 persons from remaining 7 persons as P_{1} is selected and P_{4} and P_{5} are already removed.

Number of ways = Selecting 4 persons from remaining 7 persons

= ^{7}C_{4}

By using the formula,

^{n}C_{r} =\( \dfrac{ n!}{r!(n – r)!}\)

^{7}C_{4} =\(\dfrac{ 7! }{ 4!(7 – 4)!}\)

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

= 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.

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