schumi_gmat wrote:I Think Ian we have to multiply by 5!. My solution assuming that first didgit can be 0 in an ID.
no of ways to select 5 digit number is 10^5
no of ways to select digit with 3 2's is 9*9 = 81
the 2's and other two digits can be arranged among each other in 5! ways.
Hence the digits with 3 2's = 81 * 120
Probability = 81*120/ 10^5
What is an OA?
As Ian pointed out, if you use simple factiorials, you're counting identical arrangements multiple times.
For example, let's look at the two codes:
22214
and
22214
Notice anything about those two codes? Indeed, they're identical. However, using your method, we'd have counted them as two separate items.
If we think of the three "2"s as "2a" "2b" and "2c", the two arrangements we see above could be:
(2a)(2b)(2c)14
and
(2b)(2a)(2c)14
or a number of other permutations of our 2s.
Again, as Ian stated (he's usually right, you know!), we really want to use the combinations formula, because what we're really counting is the number of different subsets of 3 items we can make out of a total of 5 items: 5C3
nCk = n!/k!(n-k)! = 5!/3!2! = 5*4*3*2/3*2*2 = 20/2 = 10
So, the chance of getting exactly 3 2s in a 5 digit code (assuming the first digit can be 0) is:
5C3 * 1^3 * 9^2 / 10^5 = 810/100000 = 81/10000
and to expand it to any 3 identical digits we just multiply by 10 (since there are 10 different digits that could be triplicated) to get 81/1000.
