Math - permutations :( help?

Question

How many 6 digit numbers are possible using the following numbers

a)111234
b)133445
c)666678

Answer 1

First, I'll do a shorter example:

1123

Normally, there are 4! = 4×3×2×1 = 24 ways to arrange four things. But that is not the case, because there are two things that are identical (indistinguishable). Pretend that one of the 1s is called "A" and the other is called "B":

AB23 and BA23 are the same (1123)
A2B3 and B2A3 are the same (1213)
A23B and B23A are the same (1231)
etc...

So there are pairs of them that are equivalent, and the reason is that we can rearrange A and B as much as we want. For two things (A and B), that means we divide by 2!.

There are 6 things in each of your problems.

a) Here, we have 6! ways to arrange them, but the 1s have to be indistinguishable. So we divide by the number of ways to arrange the 1s (since such arrangements are the same). 3! ways to do that.

Answer: 6!/3! = 120

b) Here, we have 2! ways to rearrange the 3s, and 2! ways to rearrange the 4s. That means our answer is:

6! / (2! 2!) = 180

c) Here, we have four 6s. So we want:

6! / 4! = 30

For more on this, see:
http://en.wikipedia.org/wiki/Multinomial_theorem

Answer 2

number of required digits factorial n!
divided by number of times of every digit factorial.

a. 6! / (3! 1! 1! 1!) 3 ones 1 two 1 three and 1 four
=6*5*4*3*2*1/(3*2*1*1*1*1)
=120
b. 6! / ( 1! 2! 2! 1!) 1 one 2 threes 2 fours 1 five
=6*5*4*3*2*1/(1*2*1*2*1*1)
=180
c. 6! / ( 4! 1! 1!) 4 sixes 1 seven 1 eight
=6*5*4*3*2*1/(4*3*2*1*1*1)
=30