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

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) 4P2× 7P2 (B) 4P2× 7C2 (C) 4C2× 7C2 (D) 4C2× 7P2

Tardigrade Admin · Accepted Answer

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.
Solution

We have,

x_{1} + x_{2} + x_{3} = 6,

where

x_{1}, x_{3} \geq 0 an d x_{2} \geq 1

\Rightarrow x_{1} + (x_{2} + 1) + x_{3} = 6, w h ere x_{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}

Q. 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

$^{4} P_{2} \times^{7} P_{2}$

$^{4} P_{2} \times^{7} C_{2}$

$^{4} C_{2} \times^{7} C_{2}$

$^{4} C_{2} \times^{7} P_{2}$

Q. 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

4P2​×7P2​

4P2​×7C2​

4C2​×7C2​

4C2​×7P2​

$^{4} P_{2} \times^{7} P_{2}$

$^{4} P_{2} \times^{7} C_{2}$

$^{4} C_{2} \times^{7} C_{2}$

$^{4} C_{2} \times^{7} P_{2}$