# Combination problem

• May 18th 2008, 08:38 AM
AshleyT
Combination problem
Eight cards are selected with replacement from a pack of 52 playing cards, with 12 pictures, 20 odd cards and 20 even cards.

- How many different sequences of 8 are available?

Thanks x
• May 18th 2008, 10:27 AM
Soroban
Hello, AshleyT!

Could you supply the exact wording of the problem?

And the entire problem?
We don't need the breakdown of the types of cards.
. . So I suspect there are more parts to the question.

Quote:

Eight cards are selected with replacement from a pack of 52 playing cards.
How many different sequences of 8 are available?

Since there are 52 different (identifiable) cards,

. . there are: .$\displaystyle 52^8 \:\approx\:5.346 \times 10^{13}$ possible sequences.

• May 18th 2008, 11:34 AM
AshleyT
Quote:

Originally Posted by Soroban
Hello, AshleyT!

Could you supply the exact wording of the problem?

And the entire problem?
We don't need the breakdown of the types of cards.
. . So I suspect there are more parts to the question.

Since there are 52 different (identifiable) cards,

. . there are: .$\displaystyle 52^8 \:\approx\:5.346 \times 10^{13}$ possible sequences.

Hey, thanks very much for the reply :).

There are second parts of the question but that was the first part, and the part i was stuck with. I'm going to attempt the other parts now.

Basically the chapter is based around factorials and holds no examples of using powers, so i didn't think about that.

At first i was thinking it would be 52Cr8 * something.

Is there any chance you could explain why it is 52^8 please?

Thankyou :).
• May 18th 2008, 12:03 PM
Plato
Quote:

Originally Posted by AshleyT
Eight cards are selected with replacement from a pack of 52 playing cards, with 12 pictures, 20 odd cards and 20 even cards. How many different sequences of 8 are available?

Are you sure it said “with replacement”?
If it did then the same card can be drawn several times, as many as eight.
Thus the answer $\displaystyle 52^8$.

Please check to see if it could be “without replacement”.
If it does, also see if it does say “sequences” instead of “hands”.
• May 18th 2008, 01:59 PM
AshleyT
Quote:

Originally Posted by Plato
Are you sure it said “with replacement”?
If it did then the same card can be drawn several times, as many as eight.
Thus the answer $\displaystyle 52^8$.

Please check to see if it could be “without replacement”.
If it does, also see if it does say “sequences” instead of “hands”.

I see, thankyou :).

Yea it says 'with replacement' and 'sequences'.