I am working through a problem and I would like to have some advices.
I am trying to solve counting problems without using shortcuts as binomial coefficients and such ( mainly to build intuition and be able to solve more complicated problems without getting too confused on when to use this formula or not).
I am trying to solve the following problem(I reduced a larger problem to the following toy example ) Suppose we have 3 letters, A,B,C. We Want to count the number of string containing 2 letters where A can be repeated. So basically (AA,AB,AC,BC) = 4.
I am trying to find a systematic way of solving these problems. So I start like this :
Consider our alphabet of A1, A2, B, C. The number of ordered string containing 2 characters is 4*3=12
Now the number of string containing 2 characters where order doesn’t matter is 4*3/2!=6.
From there I do not know how to modify the formula to eliminate the redundant As. Since all multiplication or division are valid only if it applies to all strings.
Any ideas ?
Here is the larger problem:
how many ordered full houses (eg KKKQQ) are there in a deck of 52 cards ( holding 5 cards ). My solution :
We have 52 choices for the first card, let’s say we picked a king, we have then 3 choice to pick the next king, and 2 choices for the last. Then we have 48 choice for the card constituting the pair and it follows 3 choice for the second card of the paire.
So we have 52*3*2*48*3. This will count all full houses of the form (KKKQQ) where K could be any card and Q any card different than K.
Now we need to divide by 3!*2! To account for the permutations of KKK and QQ. Then to get all the possible ordering we need to multiply by 5! To have a full permutation.
By doing the last step we also allowed unnecessary permutations (the Ks for example) so we need to divide again by 3!*2! But according to the results I see online I do not get it right...
How to intuitively and systematically get the answer right ?
You can do like this:
First just take A,B,C and count the number of ways, $\binom 32$ = 3
Now, just count if we take 2 A's: only one way AA.
So, total number of ways = 3 + 1 = 4
If this doesn't help you, please share your bigger problem.
For the bigger problem You need 3 cards of same kind and two of another kind. Choose your first card of one kind. $\binom {13}1$ and then choose rest 3 cards in $\binom43$ . Now choose second type card in $\binom{13-1}{1}$ and 2 cards from those 4 cards $\binom{4}{2}$.
So, $\binom{13}1 \binom 43 \binom{12}1 \binom42$.