[vinvinvin]

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 sequences of 8 cards are possible?
(b) How many sequences in part (a) will contain 3 picture cards, 3 odd numbered cards and 2 even numbered cards.

Part (a) i know how to do.

52^8=5.34597... x10^13

How should i approach part B?

Thanks

[notnek]

Firstly, for each type (picture, odd, even), how many possibilities are there for each if we consider them separately?

Next, we need to work out the different ways of ordering these cards into 8 "gaps". So let's choose the odd cards first.

How many ways can you slot 3 odd cards into 8 gaps?

Then you can work out the number ways of adding the picture cards to the sequence (remember there are 5 spaces left to fill).

The two even cards will slot into the remaining spaces.

Does this help? Post some working if you're still stuck.

[ghostwalker]

IF P is picture, E even, O odd.

You could consider how many possibilities are there for a sequence of PPPOOOEE in that order.

Then, how many arrangements are there for "PPPOOOEE".

[notnek]

(Original post by ghostwalker)
IF P is picture, E even, O odd.

You could consider how many possibilities are there for a sequence of PPPOOOEE in that order.

Then, how many arrangements are there for "PPPOOOEE".

This is the same as the method I gave, no?

Finding how many arrangement there are for "PPPOOOEE" is the same as considering how to slot the P's, O's and E's into 8 gaps.

[ztibor]

(Original post by vinvinvin)
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 sequences of 8 cards are possible?
(b) How many sequences in part (a) will contain 3 picture cards, 3 odd numbered cards and 2 even numbered cards.

Part (a) i know how to do.

52^8=5.34597... x10^13

How should i approach part B?

Thanks

The number of arranging 3 picture, 3 odd and 2 even number card is or with permutation

but the number of selections of k cards from n with replacement is
e.g for picture cards k=3 n=12 so
Different 3 cards means different sequence.
Multiply these possibilities (the selections and arrangement)

[ghostwalker]

(Original post by notnek)
This is the same as the method I gave, no?

Finding how many arrangement there are for "PPPOOOEE" is the same as considering how to slot the P's, O's and E's into 8 gaps.

Yes; it's just a slightly different way of seeing it, which, from previous post on here, I thought would be more recognizable.

[ztibor]

(Original post by vinvinvin)
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 sequences of 8 cards are possible?
(b) How many sequences in part (a) will contain 3 picture cards, 3 odd numbered cards and 2 even numbered cards.

Part (a) i know how to do.

52^8=5.34597... x10^13

How should i approach part B?

Thanks

The number of arranging 3 picture, 3 odd and 2 even number card is or with permutation

but the number of selections of k cards from n with replacement is
e.g for picture cards k=3 n=12 so
Different 3 cards means different sequence. Multiply these possibilities(selection and arrangement)

[ghostwalker]

(Original post by ztibor)
but the number of selections of k cards from n with replacement is
e.g for picture cards k=3 n=12 so
Different 3 cards mean different sequence.
Multiply these possibilities.

With replacement:

I think you're assuming that the order doesn't matter, but it does since we are dealing with a sequence.

If order didn't matter, the binom term for k from n would be

[ztibor]

(Original post by ghostwalker)
With replacement:

I think you're assuming that the order doesn't matter, but it does since we are dealing with a sequence.

If order didn't matter, the binom term for k from n would be

Tanks.
Yes,I think at the first step the order does not matter, only that which cards
can I select, because different cards may give different possible sequence, then at the second step I should calculate with permutation or combination the number of arrangement and should multiply the given values.

[ztibor]

(Original post by ghostwalker)
With replacement:

I think you're assuming that the order doesn't matter, but it does since we are dealing with a sequence.

If order didn't matter, the binom term for k from n would be

Tanks.
Yes,I think at the first step the order does not matter, only that which cards
can I select, because different cards may give different possible sequence, then at the second step I should calculate with permutation or combination the number of arrangement and should multiply the given values.

[ghostwalker]

(Original post by ztibor)
Tanks.
Yes,I think at the first step the order does not matter, only that which cards
can I select, because different cards may give different possible sequence, then at the second step I should calculate with permutation or combination the number of arrangement and should multiply the given values.

Sorry, don't follow you.

I make it

Spoiler:

Show

Not sure whether you're agreeing with that or not.

[ztibor]

(Original post by ghostwalker)
Sorry, don't follow you.

I make it

Spoiler:

Show

Not sure whether you're agreeing with that or not.

I agree that, I though that the order did not matter but that was not that case

I think e.g the 12 picture cards have different picture, but I
can select the same one 3 times, or I can select the same twice
and another one, or different 3 cards.
So I have for selection f.e. 12+12*11+12*11*10 sequence of 3 selected cards
which is not equal with 12^3.
Same for selecting of odd or even number cards.
For arranging the selected cards at the 8 places only the order of group of cards (3.3.2) does matter. I agree that this would be 8!/(3!*3!*2!) or 8C3 x 5C3 x2C2.

[ghostwalker]

(Original post by ztibor)
I think e.g the 12 picture cards have different picture, but I
can select the same one 3 times, or I can select the same twice
and another one, or different 3 cards.
So I have for selection f.e. 12+12*11+12*11*10 sequence of 3 selected cards
which is not equal with 12^3.

If you have the same card twice, then with ordering it could be xxy, xyx, yxx.
So, 12*11*3, and then they do all add to 12^3.

[Vip3rgt9]

(Original post by vinvinvin)
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 sequences of 8 cards are possible?
(b) How many sequences in part (a) will contain 3 picture cards, 3 odd numbered cards and 2 even numbered cards.

Part (a) i know how to do.

52^8=5.34597... x10^13

How should i approach part B?

Thanks

Is part A correct? How come it's not 52C8 X 8!=3.0342... x10^13?

[notnek]

(Original post by Vip3rgt9)
Is part A correct? How come it's not 52C8 X 8!=3.0342... x10^13?

Yes it is correct.

The question is asking to find how many sequences of 8 cards are possible. Since there are no conditions on the cards, you can think of the 52 cards as identical.

So choice doesnt matter. i.e. a sequence of 1,2,3,4,5,6,7,8 (of any suit) is equivalent to 2,4,1,6,7,3,8,5.