The problems on this page both deal with hands taken from a deck of 24 cards:
The 9, 10, Q, K, A of each of the four suits.

a. How many 2-card hands are possible which satisfy:
The hand has exactly one Queen
AND
The hand has exactly one heart

b. How many 3-card hands are possible which satisfy:
The hand has exactly one Queen
AND
The hand has exactly one heart

The problems on this page both deal with hands taken from a deck of 24 cards:
the 9, 10, Q, K, A of each of the four suits.


a. How many 2-card hands are possible which satisfy:
the hand has exactly one Queen
AND
the hand has exactly one heart

Since the hand must have exactly one queen and exactly one heart, we can assume that the queen of hearts can not be used.

Choose 1 queen from the 3 queens allowed (not including the queen of hearts). C(3,1)

Choose 1 heart from the 5 hearts in the deck (not including the queen of hearts),
C(5,1)

5*3=15 different ways of getting exactly one heart and one queen in a 2 card draw from a 24 card deck.

For the probability, there are C(24,2)=276 different 2 card hands in a 24 card deck. Therefore, the probability of getting such a hand is
\frac{15}{276}=\frac{5}{92}

Another way, choose the queen out of the 3 which are not hearts, 3/24=1/8

Since the previous card was not replaced, choose the heart from the 5 hearts in 23 remaining cards, 5/23.

But they can be drawn in 2 different orders:

2(1/8)(5/23)=\fbox{5/92}




b. How many 3-card hands are possible which satisfy:
the hand has exactly one Queen
AND
the hand has exactly one heart

Choose the queen out of 3 queens as before, the heart out of 6 hearts, and the third card choose from the 15 left.

There are 3*6*15=270 different 3-card hands with exactly one heart and one queen drawn from a 24 card deck.

\frac{C(5,1)C(3,1)C(15,1)}{C(24,3)}=\frac{225}{202 4}

Or

6(\frac{5}{24})(\frac{3}{23})(\frac{15}{22})=\frac {225}{2024}

In any case, the problem came back incorrect for both 18 and 15.

Yes, I noticed my oversight right ater I posted, then my internet went down :(

It would seem to me that it would be as I made my changes in my previous post.

They may be counting the case where the queen of hearts is used alone for exactly one heart and one queen, then the other card drawn from the remaining 15. If so, that would be added on, not multiplied. Try 30 then.

Try 30 then. 15 for the previous cases outlined, then 15 if the queen of hearts is used. In this case, if the queen of hearts is used, then there are 15 cards left to choose from because we could not have anymore hearts or queens.

The problem with these questions, it's that it's difficult to know what they are looking for... =/

Another way:
1 Queen and 1 heart, other than Queen of hearts = 3C1 x 5C1 = 15
OR
1 Queen of Hearts, and any other card non queen, non heart = 1C1 x 15C1 = 15

P(1 Queen and 1 Heart) = (15+15)/24C2 = 30/272

Second one:
1 Queen, 1 Heart, and any other card, except Queen of hearts = 3C1 x 5C1 x 15C1 = 225
1 Queen of hearts, and 2 other cards non queen, non hearts = 1 x 15C2 = 105

P(1 Queen, 1 Heart and one other) = (225 + 105)/24C3 = 330/2024

EDIT: Typo corrected. Replaced 254 by 272 for 24C2

Yes, O agree with U Man... Interpretation.

But, there are 276 different ways to draw 2 cards from 24.

\frac{24!}{2!\cdot 22!}=\frac{24\cdot 23}{2}=12\cdot 23=276