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I'm a little confused about how to solve the following problem. Could someone possibly give me an intuitive way to think about it. Thank you very much!

Problem: How many ways are there to choose a 5 card hand such that there are 4 cards with 4 different suits, (♧︎, ♢︎, ♡, ♤) and the fifth card could be anything.

bugsyb
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  • The answers below explain it well. I'll add that the common mistake in this sort of problem is to say: pick one card from each suit, and then pick any one of the other 48 cards, for a total of 1313131348, but that counts each hand twice, since the two cards of the same suit are ordered by distinguishing which one occurs as "one from each suit" versus "one from the remaining 48", but you don't want to order them. – Ned Nov 05 '15 at 21:01

4 Answers4

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By the Pigeon Hole Principle: the fifth card must be one of those four suits also.

Count ways to select: a suit to have two cards, two cards for that suit, and one card for each of the three others.

Assuming this is a standard 52 card deck (4 suits with 13 cards each and no jokers).

Graham Kemp
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Out of 4 suits select 1 to have 2 cards: $ {4 \choose 1} = 4 $

Then out of that suit you have $13 \choose 2$ possibilities. Times ${13 \choose 1} = 13$ for each of the other suits.

So:

$$ 4 * {13 \choose 2} * 13 ^3$$

Pieter21
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  • Why wouldn't it work if we did the following: 52 * 39 * 26 * 13 * 48? I know that it doesn't work, but I don't understand why. There are 52 ways to fill the first spot, 39 to fill the second, etc... Why doesn't this work? – bugsyb Nov 05 '15 at 20:54
  • because of reordering. 2 of clubs then 2 of spades is the same as the other way around. – Pieter21 Nov 05 '15 at 20:57
  • So then couldn't we simply divide by the number of combinations of reordering(5!)? – bugsyb Nov 05 '15 at 20:59
  • because of the difference between the suit with two cards. However, if you consider picking the three suits with 1 card first, and then 2 cards from the final suit, it could be easier to understand (523926/3! = 4*13^3) – Pieter21 Nov 05 '15 at 21:06
  • Oh I get it so instead of dividing by 5!. We would divide by 4! * 2!. The 4! represents the reordering of the first 4 cards. And the 2! would represent the reordering of the two cards with the same suit. Thanks! – bugsyb Nov 05 '15 at 21:08
  • Nope, 3! to reorder the 3 suits, and 2! between the two in the same suit. That causes a factor 4. – Pieter21 Nov 05 '15 at 21:10
  • Hmm I'm not getting the right answer by that method. I'm pretty sure its 4! * 2! Because even the duplicated suit has to be reordered among the other suits. My calculation is: 5239261348/(2!*4!). This seems to get the correct answer. – bugsyb Nov 05 '15 at 21:17
  • @user3594588: You are right, as is your explanation in your comment before this. – true blue anil Nov 06 '15 at 04:41
  • The answer is indeed correct (as it is the same as I had, 685464). It is also $13^3 * 48 / 2$: take any of the 4 suits, take any of the remaining 48, divide by 2 to deduplicate the suit of the last color. – Pieter21 Nov 06 '15 at 12:35
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First card is anything
Second card 39/51
Third card is 26/50
Fourth card is 13/49
Last card is anything

10.55%

paparazzo
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The answer to this is relatively simple.

The first intuition that comes to anyone's mind is this:

Pick one card from each suit, and then pick any one of the other 48 cards, for a total of 13 * 13 * 13 * 13 * 48

However, this is wrong due to one simple reason. This includes all the cases twice. For example, if it will include both (2 ♧︎, 3 ♢︎, 3 ♡, 3 ♤, 4 ♧︎) as well as (4 ♧︎, 3 ♢︎, 3 ♡, 3 ♤, 2 ♧︎). So, we will simply divide our answer by 2.

Hence, the answer would be 13 * 13 * 13 * 13 * 48 / 2 = 685464.

This is your answer based on Principle of Counting.

Dhruv Badaya
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