# Permutation and Combination Help!

• Mar 18th 2008, 11:49 AM
looi76
Permutation and Combination Help!
Question:

Eight cards are selected with replacement from a standard pack of 52 playing cards, with 12 picture cards, 20 odd cards and 20 even cards.
(a) How many different sequences of eight cards are possible?
(b) How many of the sequences in part (a) will contain three pictures, three odd-numbered cards and two even-numbered cards?
(c) Use parts (a) and (b) to determine the probability of getting three picture cards, three odd-numbered cards and two even-numbered cards if eight cards are selected with replacement from a standard pack of 52 playing cards.

Attempt:
• Mar 18th 2008, 11:54 AM
digitalis77
a.)
a.) Start out with 52 cards taken eight at a time (52C8)
• Mar 18th 2008, 12:00 PM
looi76
Quote:

Originally Posted by digitalis77
a.) Start out with 52 cards taken eight at a time (52C8)

I did that and got $752538150$ but the answer in the text book is $5.346\times10^{13}$. I am confused!
• Mar 18th 2008, 12:02 PM
digitalis77
dont those #s match
• Mar 18th 2008, 12:03 PM
digitalis77
nevermind those #s dont match. what level math are you in?
• Mar 18th 2008, 12:05 PM
looi76
Quote:

Originally Posted by digitalis77
nevermind those #s dont match. what level math are you in?

GCE AS-Level Maths
• Mar 18th 2008, 12:21 PM
Plato
Quote:

Originally Posted by looi76
Question:

Eight cards are selected with replacement from a standard pack of 52 playing cards, with 12 picture cards, 20 odd cards and 20 even cards.
(a) How many different sequences of eight cards are possible?
(b) How many of the sequences in part (a) will contain three pictures, three odd-numbered cards and two even-numbered cards?
(c) Use parts (a) and (b) to determine the probability of getting three picture cards, three odd-numbered cards and two even-numbered cards if eight cards are selected with replacement from a standard pack of 52 playing cards.

You both missed the bold red above.
a) ${52}^8 = 53459728531456$
b) $({12}^3)({20}^3)({20}^2)(8!)
$