# Probability - Combinations and permutations

Please could someone advise if the following looks correct. I am preparing for an exam and this is the only way I can get any feedback on my work.

The last 3 digits of a number have been erased. Number looks something like 24937???. Assuming that all possibilities are equally likely, find the probabilities of the following events.

a) The missing digits are 1,2,3 (in this order)

The number of potential outcomes is 1000. Therefore probability is 1/1000

b) The set of missing digits is {0,4,7}

This set of digits will appear n! times out of all the possible outcomes.

So 6/1000

c) The missing digits are all equal to each other

There is 10 possible outcomes here = {000, 111 ...999}

Therefore 10/1000

d) Two of the digits are equal to each other but the third digit is different.

I think this is 270/1000. Logically, the 2nd and 3rd numbers can form 10 pairs of equal numbers. In order for the first number to be different this means that 90 triple combinations can be formed. We can do this operation twice more. First,by making pairs from the 1st and 3rd numbers and adding a different (not equal) 2nd digit etc. Is this correct? I am unsure what formula should be applied in terms of combinations.

e) All 3 digits are different to each other

The first digit has 10 possibilities. The 2nd has therefore 9. The 3rd has 8 possibilities.

I get 720/1000

I'll assume that you are asking for a formula where $k$ digits out of $n$ are equal to each other and the rest $n-k$ digits are also equal to each other but the two groups are different. E.g. 24242 can be an example for $n=5$ and $k=3$. In this case the answer is $90 \times {n \choose k}$ since we have to choose $k$ positions for the first group and the positions of the second group is then forced. The $90$ comes from $10 \times 9$ as you write correctly yourself. – Marek Jan 9 '13 at 12:44