Question

How many 9 digit numbers can be formed using digits 2,2,3,3,5,5,8,8 so that odd digits occupy even positions?

Answer 1

There are 4 even places and 5 odd places.

Odd digits are 3355.

Odd numbers are placed in these 4 places in 4!2!.2! ways.

Even digits are 22888.

These even digits are placed in 5 odd places in 5!3!.2! ways.

∴ No. of required 9 digit numbers= 4!2!.2!.5!3!.2!

=60

Answer 2

How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions?

What does Google know about me?

The answer I think should be 60. Here's how

The number of odd numbers we have is 4, but all 4 are not distinct ie. 3,3,5,5

Now you have to fill them in Even positions. So arrangement will be something like this.

_O_O_O_O_ where O represents odd numbers.

So the number of ways to arrange these odd numbers will be 4! But as the numbers 3,5 are repeated twice, we will have to divide by 2! Twice to avoid repition of cases. It will give me 24/(2∗2)=6 ways.

Now we place the even numbers, so our arrangement will be like this.

E_E_E_E_E where E represents even numbers,

Just as above we will get a result of 5! But 2 is repeated twice and 8, thrice. So we will have to divide by 2!3! It will give me 120/(2∗6)=10 ways.

Now as we want both these arrangements together, we multiply the number of ways each arrangement is possible ie. 6∗10=60.

So 60 such nine digit numbers are possible.

Thanks!