From 2,3,4,4,5,how many numbers can be formed such that even digits are at odd places ?

Question

Now I have 2 even positions and 2 odd positions and I have 3 even numbers 2 ,4 ,4 and 2 odd numbers so one odd position will be occupied by an even number but then even no can't be at odd position so then how to approach this question ?

Answer 1

It is a five digit no.

So, 1st,3rd,5th place can be occupied by even digit like 2,4,4

They can be arranged 3!/2!=3 ways

2nd and 4th place can be arranged by 3,5 in 2! ways

So, total 3*2!=6  numbers

Comments

But it is not mentioned that we must use all digits- number can be 1-4 digits also rt?

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It must be ..1 digit...2..3...4...5 digits numbers