# Two combinatoric problems

• Aug 21st 2010, 02:34 PM
grottvald
Two combinatoric problems
Hi I have to problems reggarding combination I really need some help with.

The first one:
In have many ways can 3 cards be drawn from an ordinary card deck (52 cards) so that all of the 3 cards have the same color (3 spades, 3 hearts, 3 diamonds or 3 cubs)?

I know that just 3 cards out of 52 can be drawn in C(52, 3) ways but how do I do it when all of these must have the same color?

The second problem is very much simpler.
12 people are going to stand in an circle. in how many ways can this be arranged?
Is 11! the correct solution? It would be 12! if 12 they where standing in a straight line, right?

Thanks.
• Aug 21st 2010, 02:42 PM
undefined
Quote:

Originally Posted by grottvald
Hi I have to problems reggarding combination I really need some help with.

The first one:
In have many ways can 3 cards be drawn from an ordinary card deck (52 cards) so that all of the 3 cards have the same color (3 spades, 3 hearts, 3 diamonds or 3 cubs)?

I know that just 3 cards out of 52 can be drawn in C(52, 3) ways but how do I do it when all of these must have the same color?

The second problem is very much simpler.
12 people are going to stand in an circle. in how many ways can this be arranged?
Is 11! the correct solution? It would be 12! if 12 they where standing in a straight line, right?

Thanks.

11! is correct.

For the first part I think you mixed up color and suit. There are only 2 colors, red and black. It's simply C(26,3) + C(26,3).
• Aug 21st 2010, 03:06 PM
grottvald
Quote:

Originally Posted by undefined
11! is correct.

For the first part I think you mixed up color and suit. There are only 2 colors, red and black. It's simply C(26,3) + C(26,3).

Yeah I really mixed it up but I mean Suit. The 3 cards must all have the same suit. English isn't my first language so sometimes I mess things up. (Headbang)

A wild guess: Is it something like C(52, 3) * C(4,3) ?
• Aug 21st 2010, 03:33 PM
undefined
Quote:

Originally Posted by grottvald
Yeah I really mixed it up but I mean Suit. The 3 cards must all have the same suit. English isn't my first language so sometimes I mess things up. (Headbang)

A wild guess: Is it something like C(52, 3) * C(4,3) ?

How many ways to pick 3 hearts? C(13, 3)
How many ways to pick 3 clubs? C(13, 3)

...

Answer is 4 * C(13, 3)

Edit: Fixed typos
• Aug 21st 2010, 04:04 PM
grottvald
Thank you so much!
• Aug 21st 2010, 04:54 PM
grottvald
I got stuck on another question of the same character:

"4 identical wonderful tasty tomatoes are to be distributed (as in divided, spread) among 3 people. In how many ways can this be arranged?"

Is this even possible to solve this one purely combinatorial? Or is it a trick question?
• Aug 21st 2010, 05:16 PM
undefined
Quote:

Originally Posted by grottvald
I got stuck on another question of the same character:

"4 identical wonderful tasty tomatoes are to be distributed (as in divided, spread) among 3 people. In how many ways can this be arranged?"

Is this even possible to solve this one purely combinatorial? Or is it a trick question?

I'm not sure what we're "supposed" to assume. Seems natural enough to assume we aren't allowed to cut any tomatoes! But in terms of,

1) Are the tomatoes identical?
2) Does each person necessarily get at least one tomato?
3) Must all tomatoes be distributed?

Assuming yes, no, and yes, I get

{0,0,4} -- multiply by 3
{0,1,3} -- multiply by 3!
{0,2,2} -- multiply by 3
{1,1,2} -- multiply by 3

This gives 15 ways.