Question

Number of six digit numbers which have $3$ digits even & $3$ digit odd, if each digit is to be used atmost once is ______ .

Answer 1

_ _ _ _ _ _ (neglecting '0' - neither odd nor even)

Given even numbers must be 2,4,6,8

odd numbers must be 1,3,5,7,9

ways of choosing three even numbers from

four no. will be ${}^4{C}_3$

$||{|}^{ly}$ choosing odd no. will be ${}^5{C}_3$

Then we get 6 no. which of 3 even and

3 odd.

$\therefore$ ways of arranging then at 6 places will be

of 6!

$\therefore$ Total no. of digit number formed

will be

${}^5{C}_3{.}^4{C}_3.6!$

$=\frac{5!}{3!2!}.\frac{4!}{3!1!}\times 6!$

$=10\times 4\times 720=28800$