Question

There are ten seats out of which four are to be occupied. The number of ways of arranging four persons on these four seats such that each person has exactly one neighbour is

• A. $^{4}P_{2}\times ^{7}P_{2}$
• B. $^{4}P_{2}\times ^{7}C_{2}$
• C. $^{4}C_{2}\times ^{7}C_{2}$
• D. $^{4}C_{2}\times ^{7}P_{2}$

Answer 1

Answer: (A)

Group of two persons can be selected in $^{4}C_{2}$ ways. Let $P_{1}P_{2}$ are together and $P_{3}P_{4}$ are together. Let $x_{1}$ be the number of seats vacant at the left end, $x_{2}$ seats are between two pairs and $x_{3}$ seats are vacant at the right end.

We have, $x_{1}+x_{2}+x_{3}=6,$ where $x_{1},x_{3}\geq 0andx_{2}\geq 1$
$\Rightarrow x_{1}+\left(x_{2} + 1\right)+x_{3}=6,wherex_{1},x_{2},x_{3}\geq 0$
$\Rightarrow x_{1}+x_{2}+x_{3}=5$
Number of ways of selecting seats $=^{n + r - 1}C_{r - 1}=^{3 + 5 - 1}C_{3 - 1}=^{7}C_{2}$
Persons can interchange their seats in $2!\times 2!$ ways
$\Rightarrow $ Required ways $=^{4}C_{2}\times ^{7}C_{2}\times 2!\times 2!=^{4}P_{2}\times ^{7}P_{2}$