How do I find number of ways to choose 10 cards from a set of 7 blue and 8 red cards If given a set of $7$ blue cards and $8$ red cards and asked to pick $10$ at random how do I figure out how many possible ways there are to select those cards? I got a bit stumped after realizing $2^{10}$ doesn't really help here and I'm not sure how to calculate it in a way that would account for the fact that you can run out of one type of card. I also would like to know how to figure out the number of ways to select $5$ red and $5$ blue cards exactly.
Thank you for any help
Edit to clarify:
The cards are identical but the order they are drawn is important to the question. For example, B B B B B R R R R R and R R R R R B B B B B would count as $2$ ways to draw the cards.
 A: First, consider the case without any restriction i.e. you can choose at most $10$ red cards. As you figured out, this can be done in $2^{10}$ ways.
Now back to our original question, we overcounted the cases when there are $10$ or $9$ red cards and the cases when there are $10$ or $9$ or $8$ blue cards. We can choose $10$ red cards in only one way. When we chose $9$ red cards, there were $10$ spaces between each consecutive red cards and we can place a blue card in each space. Similarly, we get that there are $1$ and $10$ ways when there are $9$ or $10$ blue cards. For the case when there are $8$ blue cards, we use stars and bars to see that there are $\binom{10}{2}$ ways. So the answer would be $2^{10}-2(10+1)-\binom{10}{2}$.
This also matches with Syam's answer.

A lot of controversies has happened regarding the question. But if we think about the configuration of the cards, we will see that the answer will be less than $2^{10}$. Also making the problem smaller and listing all possible configurations is a good way to check the answers.
A: Do it by listing down the possible ways.
For example, we can choose

*

*7B and 3R cards --> $(10!)$/$(7!3!)$

*6B and 4R cards --> $(10!)$/$(6!4!)$

*5B and 5R cards --> $(10!)$/$(5!5!)$- This answers the last part of the question

*4B and 6R cards -->  $(10!)$/$(4!6!)$

*3B and 7R cards --> $(10!)$/$(3!7!)$

*2B and 8R cards --> $(10!)$/$(2!8!)$
A: Your question is still a little unclear, but I suppose you mean that within the blue cards, all of them are identical and within the red cards, all of them are identical, not that all of the cards are identical (since if it was to be that all of the cards are identical, then there really is no point in the question specifying the colours  - it could simply have said $15$ identical cards).
Now, if the former is true, then when picking $10$ random cards out of the $15$, it would be best to consider cases with respect to the blue cards (you can obviously do the same with respect to the red cards). Clearly, we have 6 cases - $7$ blue and $3$ red, $6$ blue and $4$ red and so on... Thus, we can just find the number of possible arrangements for each case and add them all up.
For example, the first case would give $$\binom 7 7 \binom 8 3 \frac {10!} {7! 3!},$$ where we choose $7$ out of $7$ (i.e. all) of the blues, $3$ out of $8$ of the reds and then arrange them all in a line (i.e. $10!$), but remembering also that same-coloured balls are identical (i.e. divide by $7!3!$).
You can find the number of possible arrangements for the rest of the cases in a similar fashion and answer your second part as well.
P.S. Welcome to Math SE! If this answer has helped you, please give it an upvote and the tick :)
Edit
As noted in the long discussion under my answer and corroborated with another answer, it is a MUST to first choose how many blue and red balls you want, because in the question, we are given more blue and red balls than we need. It is not that we are differentiating between each blue and each red ball (the question has indicated that they are identical within the colours) - we are simply “throwing away” the excess of each type of ball which we do not need. Identicality is accounted for in the permutations, not when we are choosing the number of each type of ball.
A: Since controversy has arisen, I thought it best to write it as an answer.
I take it that it is settled that cards of the same color are identical, ie cards are distinguished only by color.
We consider the case of $7$ blue, $3$ red

*

*firstly assume that cards are numbered $b_1,b_2,... r_1,r_2...$ etc


*It is clear that they can be chosen and permuted in $\binom77\binom{8}310!$ ways


*But in reality, all blue cards are identical, and so are all red cards, so we remove duplication of permutations to finally get
$\binom77\binom{8}3\frac{10!}{7!3!}$ as the answer
PS:
In hypergeometric problems with red balls, blue balls, yellow balls, etc each ball is generally given a fictitious identity, which is later on erased by removing permutations within the same color.
PPS
Some time back, there were a couple of questions about "how many ways" balls of varying colors could be drawn. There was quite a lot of controversy on those, here is one example, the other was even funnier. The book answers tallied with the concept matching the method used in calculating probability.
