9 married couples are going for a roller coaster ride, having 9 rows consisting of 2 seats each, numbered from 1 to 18. Number of possible arrangements such that at least 5 wives are seated in the same row as that of their husband (in seats of each row, exactly one female and one male to be seated) is of the form $P \times 2^{9} \times 9!,$ then

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sum of all the digit of P is 16

number of divisors of P is 9

sum of all he divisors of P is 1456

number of divisors of P is 6

answer is A, D.

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Detailed Solution

$P o s s i b l e w a y s o f a r r a n g e m e n t (5 c o u p l e s t o g e t h e r a n d 4 n o t t o g e t h e r) + (6 c o u p l e s t o g e t h e r a n d 3 n o t t o g e t h e r) + (7 c o u p l e s t o g e t h e r a n d 2 n o t t o g e t h e r) + (9 c o u p l e s t o g e t h e r a n d 0 n o t t o g e t h e r)$

$\begin{aligned} = (^{9} C_{5} \times 9 \times 2^{9} \times 9!) + (^{9} C_{6} \times 9 \times 2^{9} \times 9!) + (^{9} C_{7} \times 9 \times 2^{9} \times 9!) + (^{9} C_{9} \times 9 \times 2^{9} \times 9!) = 1339 \times 2^{9} \times 9! \\ P = 1339 \end{aligned}$