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Home » 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.
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.
CBSE | Class 11 | Combinations | Exercise 17.3 | Maths | RD Sharma
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.

Given:

Total persons

    \[=\text{ }10\]

Number of persons to be selected

    \[=\text{ }5\text{ }from\text{ }10\]

persons

    \[({{P}_{1}},\text{ }{{P}_{2}},\text{ }{{P}_{3}}~\ldots \text{ }{{P}_{10}})\]

It is also told that

    \[{{P}_{1}}\]

 should be present and

    \[{{P}_{4}}~and\text{ }{{P}_{5}}~\]

should not be present.

We have to choose

    \[4\]

persons from remaining

    \[7\]

persons as

    \[{{P}_{1}}~\]

is selected and

    \[{{P}_{4}}~and\text{ }{{P}_{5}}\]

 are already removed.

Number of ways = Selecting

    \[4\]

persons from remaining

    \[7\]

persons

    \[={{~}^{7}}{{C}_{4}}\]

By using the formula,

    \[^{n}{{C}_{r}}~=\text{ }n!/r!\left( n\text{ }-\text{ }r \right)!\]

    \[^{7}{{C}_{4}}~=\text{ }7!\text{ }/\text{ }4!\left( 7\text{ }-\text{ }4 \right)!\]

Or,

    \[=\text{ }7!\text{ }/\text{ }\left( 4!\text{ }3! \right)\]

    \[=\text{ }\left[ 7\times 6\times 5\times 4! \right]\text{ }/\text{ }\left( 4!\text{ }3! \right)\]

Or,

    \[=\text{ }\left[ 7\times 6\times 5 \right]\text{ }/\text{ }\left( 3\times 2\times 1 \right)\]

    \[=\text{ }7\times 5\]

So,

    \[=\text{ }35\]

Now we need to arrange the chosen

    \[5\]

people. Since

    \[1\]

person differs from other.

    \[35\text{ }\times \text{ }5!\text{ }=\text{ }35\text{ }\times \text{ }\left( 5\times 4\times 3\times 2\times 1 \right)\]

    \[=\text{ }4200\]

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

    \[4200.\]

i

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