How many ways can the two cards from the suit be chosen?

How many ways can the two cards from the suit be chosen?

How many ways can the two cards from the suit be chosen?
How many ways can the two cards from the suit be chosen?

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How many ways can the two cards from the suit be chosen?

Oregon State University

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Determine whether each situation involves a permutation or a combination. Then find the number of possibilities. How many ways can a hand of five cards consisting of three cards from one suit and two cards from another suit be drawn from a standard deck of cards?

How many ways can the two cards from the suit be chosen?

Get the answer to your homework problem.

Try Numerade free for 7 days

Oregon State University

We don’t have your requested question, but here is a suggested video that might help.

Determine whether each situation involves a permutation or a combination. Then find the number of possibilities. How many ways can a hand of five cards consisting of three cards from one suit and two cards from another suit be drawn from a standard deck of cards?

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So, I am having a problem with this in that the method I use gives two completely separate answers.

Two cards are selected from a deck of $52$ playing cards. What is the probability they constitute a pair (that is, that they are of the same denomination)?

So, for the first method I reason this.

The first card picked has a $13/52$ chance of being in some suit. The second card picked has probability $12/51$ of being in the same suit.

So... The probability should be $(13/52)(12/52) = 3/52$.

The other method is by combinatorics.

I have $52 \cdot 51$ one ways of creating a pair of cards. But I have $13 \cdot 12$ different ways of creating a pair of the same suit. Now to me, the logical thing to do is to multiply this number by $4$, because I would have to count each valid pair from each suit.

This would give me

$$\frac{4 \cdot 12 \cdot 13}{52 \cdot 51}$$

What's wrong with the reasoning on the second one?