If 5 cards are deall from a standard 52 -card deck, how many different ways can three diamonds and two non-dianonds be deal?
There are five-card hands consisting of three diamonds and two non-diamonds.
(Type a whole number.)

Question

If 5 cards are deall from a standard 52 -card deck, how many different ways can three diamonds and two non-dianonds be deal?
There are

five-card hands consisting of three diamonds and two non-diamonds.
(Type a whole number.)

UpStudy AI · Accepted Answer

Let $$ D $$ be the set of diamonds in the deck and $$ N $$ be the set of non-diamonds. There are 13 diamonds and $$ 52 - 13 = 39 $$ non-diamonds.

1. The number of ways to choose 3 diamonds from 13 is given by
   $$
   \binom{13}{3} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286.
   $$

2. The number of ways to choose 2 non-diamonds from 39 is
   $$
   \binom{39}{2} = \frac{39 \times 38}{2} = 741.
   $$

3. By the multiplication principle, the total number of five-card hands consisting of three diamonds and two non-diamonds is
   $$
   \binom{13}{3} \times \binom{39}{2} = 286 \times 741 = 211926.
   $$

Thus, there are $$ 211926 $$ five-card hands consisting of three diamonds and two non-diamonds.
There are 211,926 different ways to deal three diamonds and two non-diamonds from a standard 52-card deck.

If 5 cards are deall from a standard 52 -card deck, how many different ways can three diamonds and two non-dianonds be deal?
There are five-card hands consisting of three diamonds and two non-diamonds.
(Type a whole number.)

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If 5 cards are deall from a standard 52 -card deck, how many different ways can three diamonds and two non-dianonds be deal? There are five-card hands consisting of three diamonds and two non-diamonds. (Type a whole number.)

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The Deep Dive

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Latest Probability Questions

If 5 cards are deall from a standard 52 -card deck, how many different ways can three diamonds and two non-dianonds be deal?
There are five-card hands consisting of three diamonds and two non-diamonds.
(Type a whole number.)