### Solutions to the second test

1. (28 points) This problem is about license plates consisting of 3 letters followed by 3 numbers,

2. for instance ABC 123.
1. Find the number of license plates which have a triple letter.
(So, AAA 012 would be a particular instance.)

Solution: 26·1·1·10·10·10 = 26,000

2. Find the number of license plates which have a triple number.
(So, XYZ 999 would be a particular instance.)

Solution: 26·26·26·10·1·1 = 175,760

3. Find the number of license plates which have a triple letter anda triple number.

4.
Solution: 26·1·1·10·1·1 = 260

5. Find the number of license plates which have a triple letter ora triple number.

Solution: 26,000 + 175,760 - 260 = 201,500

3. (14 points) Expand by using the binomial theorem:

Solution: (x - 2y)5 = 20x5y0 - 5·2x4y1 · 22x3y2 · 23x2y3 + 5·24x1y4 - 25x0y5,

so that (x - 2y)5 = x5 - 10x4y + 40x3y2 - 80x2y3 + 80xy4 - 32y5.

4. (16 points) A certain club has 20 members.
1. In how many ways is it possible to nominate 4 members for the offices of President, Vice President, Secretary, and Treasurer (in that order ... ) ?

Solution: 20·19·18·17 = 116,280

2. In how many ways is it possible to select a committee consisting of 4 members of the club?

Solution: = 4845

5. (14 points)
Two pairin a five card poker hand has two of one kind, two of another kind, and the fifth card of a third kind, for instance 2-2-Q-Q-A. Determine the number of two pair hands.

Suggestion: Check to be sure that your answer doesn't count anything more than once!

Method 1:· {·· 3 ··· 3}· 44 = 2808 · 44 = 123,552

Method 2:· {13 ·· 12 · }· 44 = 2808 · 44 = 123,552

Remark: the "extra" factor of  outside the brackets is to eliminate duplicates because the two pairs in a given hand could have been chosen in two different orders. For instance, we should not count 2-2-Q-Q-A and also Q-Q-2-2-A as being different hands!

Another way to see this. If we just do this calculation:
{ ·· 3} ·{ · · 3} · 44 ,
then we are counting the ways to do the following steps:
• choose one pair;
• choose another pair (with a different numerical value);
• choose a fifth card (with a still different numerical value).
So, if we have chosen two given pairs in a particular order, we can get the same hand by choosing the same two pairs in the the opposite order. To eliminate the duplication, we have to multiply by 1/2.

6. (14 points) How many rearrangements of the letters of the word ALABAMA are there?

Solution:· 3! = 210
(We first choose places for the four A's, and then distribute the other 3 letters among the remaining places.)

7. (14 points) In how many ways can 10 identical coins be distributed to 3 people?

Suggestion: This can be considered as counting solutions of x1 + x2 + ... + xn = k.
In that case, one has to determine the correct values for nand k.

Solution: We take n= 3 and k= 10. So, the number of solutions is ,
or  = 66.

For questions: roberts@math.umn.edu

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