| A Counting Problem |
Draw six cards from a standard deck of 52 playing cards without replacement. How many distinct ways can you choose these six cards so that:
1) The first card drawn is a spade,
2) the second card drawn is also a spade,
3) the third card drawn is a club,
4) the fourth card drawn is a diamond,
5) the fifth card drawn is a red card (either a diamond or a heart), and
6) the sixth and final card drawn is an ace?
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Problem Moderated by: Sasha |
| Problem Solution |
Work backwards.
The last card drawn is an ace. Divide it into four cases:
a) The ace of spades is drawn last. How many different ways could you have drawn the first five cards?
1) 12 (13-1, since you can't choose the ace of spades)
2) 11 (since you just drew a spade)
3) 13 (there are thirteen clubs left)
4) 13 (there are 13 diamonds)
5) 25 (26-1, since you chose a red card in the previous step)
So, multiply these numbers together to find the number of ways conditions 1) through 6) can be satisfied if the last card drawn is the ace of spades.
(12)(11)(13)(13)(25) = 557,700 ways
By a similar process, you should determine the following number of ways for each case:
b) Ace of clubs last: (13)(12)(12)(13)(25) = 608,400 ways
c) Ace of diamonds last: (13)(12)(13)(12)(24) = 584,064 ways
d) Ace of hearts last: (13)(12)(13)(13)(24) = 632,736 ways
Adding these four distinct cases gives the total number of ways steps 1) through 6) can be satisfied:
2,382,900 ways |
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