Permuting "123456728905"

Question

In how many ways can 123456728905be permuted such that, neither two 2's nor two 5's are adjacent to each other ?

I'm really confused how to ensure those conditions?

0 is allowed to come as the first. Consider these are simple characters, not as a whole number.

Can someone help me ?

Did you try to find all possible numbers minus all numbers that of two 2's or two 5's adjacent to each other ? — Airbag, Feb 02 '14 at 13:51

lsp · Accepted Answer · 2014-02-02T14:04:46.067

2

Hint: Calculate the number of ways in which the two $2's$ and the two $5's$ are always adjacent to each other. Subtract this from the total possible permutations.

Total possible permutations: $\frac{12!}{2!*2!}$

Now group the two $2's$ together and the two $5's$ together. So you have a total of $10$ strings now in place of $12$. So total possible permutations: $10!$

Required permutations would then be: $\frac{12!}{2!*2!} - 10!$

edited Feb 02 '14 at 14:04

answered Feb 02 '14 at 13:53

lsp

4,745

12! - 10! will be the answer in that case ? – vaidy_mit Feb 02 '14 at 13:55
Divided by (12!-10!) / (2! * 2!) ? – vaidy_mit Feb 02 '14 at 13:57
Nope. Take care of the $0$ that is present. It cannot occupy the first place anytime. – lsp Feb 02 '14 at 13:59
No,i told you that,'0' is possible ! Please read the question carefully ? Consider them as just characters ! – vaidy_mit Feb 02 '14 at 14:01
Cool sir ! :D Thanks understood :D – vaidy_mit Feb 02 '14 at 14:05
Your answer is almost correct except that $10!$ has nothing in denominator. – lsp Feb 02 '14 at 14:06
No problem. Here to help out – lsp Feb 02 '14 at 14:07
let us continue this discussion in chat – vaidy_mit Feb 02 '14 at 14:16

Permuting "123456728905"

1 Answers1