how many 5-card hands can be formed w/ exactly 3 red cards?

[brentwoodbc]

There are quite a few of these questions that I need to solve but I'll just show 1-2

from a deck of 52 cards how many different 5 card hands can be formed in each case,

with exactly 3 red cards?

with at least three red cards?

there is another question where it says " 12 face cards are removed form a deck containing 52, from these face cards 4 are chosen, how many combination's are possible that contain at least 2 red cards"

what do you do 50-12 = 40
40C4 ?
If I get help with one I think I can get the rest, I dont know what to do with these because they say how many red cards you get rather than ask you for the number of possible combo's.

Thanks.

 

[brentwoodbc]

Re: combination of cards

with the second one I think Im close.

12 cards so 6 are red.
so 6!/(2!2!)+
6!/(1!3!)+
6!/(0!4!)
=330

I know 6!/2! =360 :/ hmmmmm.

 

[brentwoodbc]

Re: combination of cards

I think I got it

6c3 * 6c1 + 6c4 * 6c0 + 6c2 *6c2 = 360.

 

[pka]

Re: Delete....

There are quite a few of these questions that I need to solve but I'll just show 1-2
from a deck of 52 cards how many different 5 card hands can be formed in each case,
with exactly 3 red cards?
with at least three red cards?

with exactly 3 red cards: \(\displaystyle \binom{13}{3}\binom{39}{2}\)

with at least three red cards: \(\displaystyle \sum\limits_{k = 3}^5 \binom{13}{k}\binom{39}{5-k}\)

 

[soroban]

\(\displaystyle \text{Hello, brentwoodbc!}\)

\(\displaystyle \text{From a deck of 52 cards how many different 5-card hands can be formed:}\)

\(\displaystyle \text{(a) with exactly 3 red cards?}\)

\(\displaystyle \text{(b) with at least three red cards?}\)

\(\displaystyle \text{(a) There are 26 red cards and 26 black cards in the deck.}\)
. . . . \(\displaystyle \text{We want exactly 3 red cards and 2 black cards.}\)

. . . . \(\displaystyle \begin{array}{cc}\text{There are:} & {26\choose3} \,=\,2600\text{ ways to get 3 red cards.} \\ \\[-3mm] \text{There are:} & {26\choose2} \,=\,325\text{ ways to get 2 black cards.} \end{array}\)

\(\displaystyle \text{Therefore, there are: }\,2600\cdot325 \:=\:\boxed{845,\!000}\text{ ways to get exactly 3 red cards.}\)



\(\displaystyle \text{(b) We want: }\,\text{(5 reds) or (4 red, 1 black) or (3 red, 2 blacks)}\)

. . . .\(\displaystyle \text{There are: }\:{26\choose5} = 65,\!780\text{ ways to get 5 red cards.}\)

. . . . \(\displaystyle \text{There are: }\:{26\choose4}{26\choose1} = 388,\!700\text{ ways to get 4 red, 1 black.}\)

. . . . \(\displaystyle \text{There are: }\:{26\choose3}{26\choose2} = 845,\!000\text{ ways to get 3 red, 2 blacks.}\)

\(\displaystyle \text{Therefore: }\:65,\!780 \;+\; 388,\!700 \;+\; 845,\!000 \;=\;\boxed{1,\!299,\!480}\text{ ways to get at least 3 red cards.}\)

\(\displaystyle \text{The 12 face cards are removed from a deck.}\)

\(\displaystyle \text{From these 12 face cards, 4 are chosen.}\)

\(\displaystyle \text{How many combinations are there that contain at least 2 red cards?}\)

\(\displaystyle \text{Note: we are using }only\text{ the 12 face caeds.}\)
. . \(\displaystyle \text{As the saying goes, "We're not playing with a full deck."}\)


\(\displaystyle \text{Among the 12 face cards, 6 are red and 6 are black,}\)
. . \(\displaystyle \text{and we draw 4 of these cards.}\)

\(\displaystyle \text{We want: }\:\text{(2 reds, 2 blacks) or (3 reds, 1 black) or (4 reds)}\)


. . \(\displaystyle \text{There are: }\:{6\choose2}{6\choose2} = 225\text{ ways to get 2 reds, 2 blacks.}\)

. . \(\displaystyle \text{There are: }\:{6\choose3}{6\choose1} = 120\text{ ways to get 3 reds, 1 black.}\)

. . \(\displaystyle \text{There are: }\:{6\choose4} = 15\text{ ways to get 4 reds.}\)


\(\displaystyle \text{Therefore, there are: }\:225 + 120 + 15 \:=\:\boxed{360}\text{ ways to get at least 2 red face cards.}\)