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

  • Thread starter Thread starter nari
  • Start date Start date
nari
Messages
4
Reaction score
0
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?
 
Mathematics news on Phys.org
nari said:
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?
 
Jameson said:
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 :)
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top