Question
The number of numbers of 9 different nonzero digits such that all the digits in the first four places are less than the digit in the middle and all the digits in the last four places are greater than that in the middle is
A.
$$2\left( {4!} \right)$$
B.
$${\left( {4!} \right)^2}$$
C.
$${8!}$$
D.
None of these
Answer :
$${\left( {4!} \right)^2}$$
Solution :
$$\frac{{1,2,3,4}}{{ \times \times \times \times }}\,\,5\,\,\frac{{6,7,8,9}}{{ \times \times \times \times }}$$ The required number of numbers $$ = {\,^4}{P_4} \times {\,^4}{P_4}.$$