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# Probability of the game Set. (Search it up if you don't know what it is!)

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(a) How many cards are in a complete deck of Set?

(b) How many unique sets are there?

When three cards form a set, we can also count the number of attributes for which all three cards are the same. For example, in the set below, all three cards have the same color, and the same shape (but they have different numbers of shapes and different shading). (c) Find the number of sets where all three cards are the same for exactly 0 attributes.

(d) Find the number of sets where all three cards are the same for exactly 1 attribute.

(e) Find the number of sets where all three cards are the same for exactly 2 attributes.

(f) Find the number of sets where all three cards are the same for exactly 3 attributes.

__________end of question__________

What I got:

for a, I got 81 cards. I'm pretty sure that's correct.

for b, there can be 1 attribute that is different and the other 3 can be the same, there can be 2 that are different and 2 can be the same, there can be 3 that are different and 1 is the same, or all 4 attributes are different.

If 1 is different and the other 3 are the same, there are \$\binom{4}{1} = 4\$ sets. If there are 2 different and 2 are the same, there are \$\binom{4}{2} = 6\$ sets. If there are 3 different and 1 is the same, there are \$\binom{4}{2} = 4\$ sets.

But I don't know how many sets there are if all 4 are different.

To get my final answer for b, I would add all the sets up.

As for c, d, e, and f, I'm pretty sure I can just refer to b.

I'm not too sure about if my current answers for b are correct or not. Could someone check my logic?

Thanks! I would really appreciate it if you helped me!

Dec 26, 2020

### 1+0 Answers

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(b) There are 108/3 = 36 unique sets.

(c) For the first card, there are 3 ways to choose the number.  For the second, there 2 choices, and then there is only 1 choice.  So there are 3*2*1 = 6 ways that the numbers can be chosen.  Doing this for the other attributes, we get 6*6*24*6 ways.  But the order of cards doesn't matter in a set, so we divide by 3!: 6*6*24*6/3! = 864.  So there are 864 sets for part (c).

(d) The cards can have the same number, color, shape, or shading.  If all the colors are the same, then there are 6*24*6/3! = 144 sets.  If all the numbers are the same, then there are 144 sets.  We get the same number for shape and shading, so there are 4*144 = 576 sets for part (d).

(e) We need to choose two of the attributes.  There are C(4,2) = 6 ways of choosing two attributes.  For each of these two attributes, there are 3 options.  For the other two attributes, there are 3 ways of assigning the choices, so there are 6*3*3*3*3 = 486 sets for part (e)

(f) First we choose which attributes are the same.  There are C(4,3) = 4 ways of choosing which attributes are the same.  There are then 3*3*4 = 36 ways to assign which is which for each of these three attributes, and there are 4 ways to assign the choices for the fourth attribute, so there are 4*36*4 = 576 sets for part (f).

Jan 8, 2021