There are 3 basic categories here, as there has to be at least one of one number, and two of the other two numbers. So we have:


So I take each these and permutate. Which gives me $3 \times 5!$ which is wrong. The answer is $90$. I know it is wrong because $3 \times 5! > 3^5$ which shouldn't be possible. I can't find any intuitive reason why my answer is wrong.
How did I approach it wrong?

  • 2
    $\begingroup$ There aren't $5!$ permutations of, say, $01122$. $\endgroup$ – lulu Jun 20 '19 at 0:33

There are some repeated bits (like $1$'s and $2$'s in the first category). So you can't permute them with $5!$ since swapping $1$'s or $2$'s in the first category doesn't change the string, similar in the second and the third category. Therefore, your answer should be $$3\cdot\frac{5!}{2!2!} = 90$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.