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# Combinatorics/Permutations - Stuck On This Q! :-( Watch

1. 8 Cards selected with replacement from a pack of 52. 12 picture cards, 20 odd cards, 20 even cards.

a) How many different sequences of 8 cards are possible?

I did this right: 52 choices in card 1, 52 in card 2 ... 52 in card 8
=> No. Sequences = 52^8

b) How many of these sequences contain 3 picture cards, 3 odd cards and 2 even cards?
I've had many goes at this, but can't get it right.

2. Probability of getting such a set multiplied by the degeneracy of that set...tried that?

ie, if I have 2 children, what are the odds that I have a son and a daughter?

well the possible combinations are: ss, sd, ds, dd....so (1/4)*2 = 1/2
3. seem to have misread the question, but concept is still the same:

think about the simplest case of a sequence of 8 cards:

pic,pic,pic,odd,odd,odd,even,eve n

how many of these are there?

12*12*12*20*20*20*20*20.

now how many unique permutations of p,p,p,o,o,o,e,e are there? once you figured this out it should become clear
4. (Original post by Plutoniummatt)
Probability of getting such a set multiplied by the degeneracy of that set...tried that?
Probability of such a set multiplied by 52^8 should give me the answer, if that's what you mean?

(Original post by Plutoniummatt)
ie, if I have 2 children, what are the odds that I have a son and a daughter?

well the possible combinations are: ss, sd, ds, dd....so (1/4)*2 = 1/2
If I have a sequence of 8, what are the odds of 3 pic, 3 odd, 2 even?

8! permutations for 8 distinct cards
(8!)/(3!3!2!) permutations for 3 pic, 3 odd, 2 even.
So P = 1/(3!3!2!) = 1/72
Then (1/72)*52^8 should give me the number of sequences overall.

But I don't think this is right, and the book disagrees with me too.

Imagine there are 8 cards. 3 even, 3 odd, 2 pictures. Imagine also that all even, odd and picture cards are the same. Then the number of ways of arranging the different cards is 8C3 *5c3*2C2.

but each picture, odd or even card is distinct. Therefore we multiply the number we just worked out by
20^3*12^3*20^2.

This is because 3 cards are 1 of 20 different cards, 2 cards are 1/20 different even cards and 3 cards are 1/12 different odd cards.

I hope this is right.
6. what answer do you get? Im abit rusty on this but i seem to get:

(8!)/(3!3!2!) multiplied by (20^5)*(12^3)

which is about 3.09 x 10^12

(which agrees with above poster)

explanation:

number of ways of arranging p,p,p,o,o,o,e,e is (8!)/(3!3!2!) I hope you get that.

for each unique way of arranging the above combination, there are (20^5)*(12^3) ways of making them with the cards provided, multiply them together and u get the answer.
7. (Original post by Plutoniummatt)
think about the simplest case of a sequence of 8 cards:

pic,pic,pic,odd,odd,odd,even,eve n

how many of these are there?

12*12*12*20*20*20*20*20.
Yep

(Original post by Plutoniummatt)
now how many unique permutations of p,p,p,o,o,o,e,e are there? once you figured this out it should become clear
12*12*12*20*20*20*20*20 possible sequences for the case pppooee (in that order)
8!/(3!3!2!) distinct arrangements of pppooee
So Total Number of Sequences (3p, 3o, 2e) = (12^3.20^5).(8! / 3!3!2!)
And the book agrees.
8. (Original post by Plutoniummatt)
what answer do you get? Im abit rusty on this but i seem to get:

(8!)/(3!3!2!) multiplied by (20^5)*(12^3)
Yep! Book agrees

(Original post by Plutoniummatt)
explanation:

number of ways of arranging p,p,p,o,o,o,e,e is (8!)/(3!3!2!) I hope you get that.

for each unique way of arranging the above combination, there are (20^5)*(12^3) ways of making them with the cards provided, multiply them together and u get the answer.
Yep. I get it. Thanks for all your help.
9. too easy next :P
10. (Original post by methusaleh)
too easy next :P
lol. It's easy when you can do it. Unfortunately for me, I'm a little inconsistent sometimes. I can crack hard problems and also trip up on some easy ones.

11. (Original post by Physics Enemy)
lol. It's easy when you can do it. Unfortunately for me, I'm a little inconsistent sometimes. I can crack hard problems and also trip up on some easy ones.

Thanks i just wanted some acknowledgment really lol.

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