Question 2

# How many ten-digit numbers can be formed using all the digits of 2435753228 such that odd digits appear only in even places?

Solution

we have : 2435753228
Now here we have 5 odd digits and 5 even digits and total 10 places
so 5,5,3,3,7 will arrange in 5 places and 2,2,2,8,4 will arrange in 5 places
we get : total arrangements =$$\frac{5!}{3!}\times\ \frac{5!}{2!\times\ 2!}$$
we get  $$\frac{\left(5!\right)^2}{3!\ \left(2!\right)^2}$$