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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 in each arrangement P1 must occur whereas P4 and P5 do not occur. Find the number of such possible arrangements. [Hint: Required number of arrangement =7C4× 5!]
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. [Hint: Required number of arrangement =7C4× 5!]
CBSE | Class 11 | Maths | NCERT Exemplar | Permutations and Combinations
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. [Hint: Required number of arrangement =7C4× 5!]

Solution:

It is known that

\begin{array}{l} { }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \\ =\frac{\mathrm{n} !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !} \end{array}

As per the question,

There are a total of 10 person named \mathrm{P} 1, \mathrm{P} 2, \mathrm{P} 3, \ldots \mathrm{P} 10

The no. of ways in which \mathrm{P}_{1} can be arranged =5 !=120

The no. of ways in which the others can be arranged,

{ }^{7} C_{4}=7 ! /(4 ! 3 !)=35

As a result, the required number of arrangement =\mathrm{C}_{4}^{7} \times 5

\begin{array}{l} =35 \times 120 \\ =4200 \end{array}

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