kvcpk wrote:anirban_lax wrote:I thought the problem is pretty easy till I tried understanding the division by 3! part. I don't see where I have considered ordering in the equation when I say - 52*51C17*34C17*1.
Will appreciate if you can you please explain in a little more detail?
Also, I notice that I can choose the set of 1 card first or the set of 17 cards first. Is there any specific reason why you decided to select the set consisting of 1 card, first? Ease of simplification I assume.
Thank you!
-Anirban
Yes I did it for ease of simplification.
I can choose any set first. Thats the reason We had to divide by 3!
Hope this helps!!
Hi!
The reason that you have to divide by 3! is because if you don't, you'll end up counting the same distributions multiple times.
Let's look at a simpler question to illustrate the principle:
How many different ways can the letters ABCDEF be distributed to 3 different groups of 2?
Just using combinations, we have:
6C2 * 4C2 * 2C2 = 15 * 6 * 1 = 90
However, since we don't care about which groups the cards go to, we've created some duplicate distributions.
For example, one distribution is:
AB CD EF
Another distribution is:
CD AB EF
The way we've counted the groups, we've counted each of these as a unique distribution. However, they're identical, so we need to eliminate the ones we've counted multiple times.
Since there are 3 items, there are 3! different ways to arrange them. Accordingly, to count the true number of unique distributions, we need to divide our sum by 3!.
So, the correct answer would be 90/3! = 90/6 = 15.
Similarly, in the question in this thread, we have:
(block of 17 "A") (block of 17 "B") (block of 17 "C").
We've counted:
A B C
A C B
B A C
B C A
C A B
C B A
as 6 different distributions of cards; however, all 6 of those are identical distributions. To avoid all those duplicates, we must divide our sum by 3!.
