How many ways can you draw 3 of one suit and 2 of another suit?

[nari]

Hi! I am completing a practice exam and there is this answer to a question that I do not understand. It says:

An experiment consists of drawing a hand of 5 cards from a standard deck of 52 cards. How many ways can you draw 3 of one suit and 2 of another suit?

I thought it was C(13, 3) x C(13,2). But it turns out it is C(13, 3) x C(13,2) x C(4,2). I don't understand where that last bit comes from. I know there are 4 suits, so I am guessing that bit is choosing two suits out of the four, but why is that not redundant when considering the first two parts of the answer?

 

[Jameson]

Hi! I am completing a practice exam and there is this answer to a question that I do not understand. It says:

An experiment consists of drawing a hand of 5 cards from a standard deck of 52 cards. How many ways can you draw 3 of one suit and 2 of another suit?

I thought it was C(13, 3) x C(13,2). But it turns out it is C(13, 3) x C(13,2) x C(4,2). I don't understand where that last bit comes from. I know there are 4 suits, so I am guessing that bit is choosing two suits out of the four, but why is that not redundant when considering the first two parts of the answer?

Hi nari,

Welcome to MHB. :) This is a good question!

The first two parts, C(13, 3) x C(13,2), deal with choosing select cards from a pool in the appropriate way, but say nothing about the suit. For the first term, C(13, 3), this could be any of the 4 suits right? And then for the second expression, C(13,2), this could be from any of the 3 remaining suits. We haven't dealt with the problem of where the 13's come from yet. We are counting 3H2S the same as 3S2D right now.

That's where this last piece comes in. Does that make sense or help?

 

[nari]

Hi nari,

Welcome to MHB. :) This is a good question!

The first two parts, C(13, 3) x C(13,2), deal with choosing select cards from a pool in the appropriate way, but say nothing about the suit. For the first term, C(13, 3), this could be any of the 4 suits right? And then for the second expression, C(13,2), this could be from any of the 3 remaining suits. We haven't dealt with the problem of where the 13's come from yet. We are counting 3H2S the same as 3S2D right now.

That's where this last piece comes in. Does that make sense or help?

Yes, it does make sense! Thank you :)