Question
Total number of 6-digit numbers in which all the odd digits and only odd digits appear, is
A.
$$\frac{5}{2}\left( {6!} \right)$$
B.
$$6!$$
C.
$$\frac{1}{2}\left( {6!} \right)$$
D.
None of these
Answer :
$$\frac{5}{2}\left( {6!} \right)$$
Solution :
Clearly, one of the odd digits 1, 3, 5, 7, 9 will be repeated.
The number of selections of the sixth digit $$ = {\,^5}{C_1} = 5.$$
∴ the required number of numbers $$ = \,5 \times \frac{{6!}}{2}.$$