Let $P_{n}$ denotes the number of ways of selecting 3 people out of $n$ sitting in a row, if no two of them are consecutive and $Q_{n}$ is the corresponding figure when they are in a circle. If $P_{n} - Q_{n} = 6$ , then $n$ is equal to

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Solution

The correct option is B $10$
If none of two are consecutive then $P_{n} = {^{n - 1} C}_{3}$

and for circular we will need to substract the permissible arrangemnt in which the two people at each end of row are selected.

$\Rightarrow Q_{n} = {^{n - 1} C}_{3} - {^{n - 4} C}_{1}$

$\Rightarrow P_{n} - Q_{n} = {^{n - 4} C}_{1}$

According to question,

$P_{n} - Q_{n} = {^{n - 4} C}_{1} = 6$

$\Rightarrow n - 4 = 6$

$\Rightarrow n = 10$

Hence, the hence is $10.$

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Let Pn denotes the number of ways of selecting 3 people out of n sitting in a row, if no two of them are consecutive and Qn is the corresponding figure when they are in a circle. If Pn−Qn=6, then n is equal to

Let $P_{n}$ denotes the number of ways of selecting 3 people out of $n$ sitting in a row, if no two of them are consecutive and $Q_{n}$ is the corresponding figure when they are in a circle. If $P_{n} - Q_{n} = 6$ , then $n$ is equal to