How many combinations that have at least 2 queens are possible?

[preet]

From a deck of 52 cards, the 12 face cards are removed. From these face cards, 4 are chosen. How many combinations that have at least 2 queens are possible?

The answer is 201... I can't get here. This is what I did:
case1) 4 queens can be chosen C(4,2) ways * 10 cards can be chosen C(10,2) ways
case2) 4 queens can be chosen C(4,3) ways * 9 cards can be chosen C(9,1) ways
case3) 4 queens can be chosen C(4,4) ways * 8 cards can be chosen C(8,0) ways

case 1 = 270
case 2 = 36
case 3 = 1
270+36+1 does not equal 201

what am I doing wrong? TiA

 

[JasonRox]

From a deck of 52 cards, the 12 face cards are removed. From these face cards, 4 are chosen. How many combinations that have at least 2 queens are possible?
The answer is 201... I can't get here. This is what I did:
case1) 4 queens can be chosen C(4,2) ways * 10 cards can be chosen C(10,2) ways
case2) 4 queens can be chosen C(4,3) ways * 9 cards can be chosen C(9,1) ways
case3) 4 queens can be chosen C(4,4) ways * 8 cards can be chosen C(8,0) ways
case 1 = 270
case 2 = 36
case 3 = 1
270+36+1 does not equal 201
what am I doing wrong? TiA

Here is your mistake.

I'll show for Case 1, then you can solve the rest.

4 Queens can be chosen, which is 4 C 2 ways. The problem is here. You said that 10 cards can be chosen, but that means a Queen can be chosen. That's wrong you can't choose a Queen again. So, it is 2 cards can be chosen from 8, which is 8 C 2.

See where this is going?

Now, solve it.

 

[Tide]

HINT: If exactly two queens are drawn, how many non-queens are available to complete the foursome?

 

[forevergone]

If you do this, It's better to do it the reverse way. Find the universal amount of cards and subtract the complement, which is Tide's method: how many non-queens are available to complete the foursome. Most questions like these, if you take a backwards approach to it, it's a lot easier.

Cases can get a bit confusing.

So:

n(u) = 12C4

Let A be the # of sequences that have only 1 queen

n(a) = 9C4

Let B be the # of sequences that have no queen

n(b) = 8C4

Total number of combinations with at least 2 queens = n(u) - n(a) - n(b)
= 12C4 - 9C4 - 8C4

 

[preet]

Thanks ... I feel pretty stupid making a mistake like that.

...forevergone, your cases don't work out... I get an answer of 299 =p

 

[forevergone]

It's probably a mathematical error, but the approach should be like that. If you use the complement and subtract it from the universe, you should get the right answer.