How many positive integers can be formed using all the digits 6, 3, 5

ashishpathak

We have 3,3,3,5,5 to put into the five odd-numbered slots and 2,4,6,6 to put into the four even-numbered slots.

Odds:
If we know what we're doing, we could use \(\frac{5!}{3!2!} = \frac{5*4}{2} = 10\).
But even if we don't, it's probably just as fast to simply count where the 5s could go (the 3s go in the other slots). The 5s could go in slots 1&3, 1&5, 1&7, 1&9, 3&5, 3&7, 3&9, 5&7, 5&9, 7&9. That's 10 ways for the odds.

Evens:
If we know what we're doing, we could use \(\frac{4!}{2!} = 4*3 = 12\).
But even if we don't, we can reason that there are 4*3*2*1 ways to place four things into the four even slots, and then we need to divide by 2 since the 6s are interchangeable, so 4*3 = 12 ways for the evens.

Multiply those together and we have 10*12 = 120.

Answer choice B.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.