Illinois State University Mathematics Department
MAT 312: Probability and Statistics for Middle School Teachers Dr. Roger Day (day@ilstu.edu) 
Possible Solutions to Counting Problems 
1. 
4! 

2. 
4*4*4*4=4^4 

3. 
P(14,5) 

4. 
C(12,4) 

5. 
There are (8*7)/2=28 ways for the 8 teams to play one game with every other team, so we double that to get 56 games in an 8team conference where every team plays every other team twice. 

6. 
For a multistate lottery, five numbers are to be picked from 44 different numbers, with no repetition. Assuming that order does not matter, how many different 5number selections are possible? C(44,5) 

7. 
A computer program selects three numbers from the set {1,2,...,20}, with repetition allowed. (a) How many different 3number selections can be made? (b) How many of these are sets with no value greater than 5? (a) 20*20*20=20^3=8000 

8. 
A field of 12 horses will run the Dentchfield Sweepstakes next week in Millbank. Assuming there are no ties, how many different winplaceshow finishes (firstsecondthird places) are possible? P(12,3) 

9. 
At one of the backstage Oscars parties Sunday night, exactly 55 handshakes took place, where everyone in attendance shook hands with everyone else. How many people were at this party? Let n represent the number of people present. then n(n1)/2 = 55 (see problem 5). The value n=11 satisfies the equation. There were 11 people present. 

10. 
A ballot on the next county election has 6 referenda on it. If a voter can vote YES, NO, or choose not to vote on each issue, how many different ways can a referenda ballot be marked? 3*3*3*3*3*3=3^6 

11. 
Suppose that a fair die is rolled seven times and the result on the faceup side is recorded. How many different ways are there for the seven rolls to result in two 1s, three 2s, and two 6s? Is this a vert likely outcome overall? We want the record to show the following values: 1,1,2,2,2,6,6. How many arrangements are there for the seven values? If they were all different, there would be 7! arrangements. however, there is duplication we need to adjust for. We divide by (2!3!2!) to account for ways the duplicates can be rearranged without changing the actual arrangement of the seven items. Therefore, there are 7!/(2!3!2!) or 210 ways to get the desired outcome. Is this very likely? There are 6*6*6*6*6*6*6=6^7=279,936 different results possible when seven dice are rolled. Comparing 210 to this, our result represents approximately 0.00075 of all possible results. It is quite unlikely! 

12. 
A seafaring ship carries 10 different signal flags. A specific message is sent by choosing and arranging from two to four different flags. How many different messages can be sent under these conditions? When two flags are used, there are P(10,2) possible messages. When three flags are used, there are P(10,3) possible messages. When four flags are used, there are P(10,4) possible messages. Because two, three, and fourflag messages have nothing in common, we add the results. There are P(10,2)+P(10,3)+P(10,4) different messages possible under these conditions. 

13. 
The head table at an annual awards banquet is to have eight speakers seated in a line. (a) How many arrangements are there for those seated at the head table? (b) Suppose the eight were seated at a circular table with eight chairs. Now how many seating arrangements exist? (a) P(8,8)=8! 

14. 
One of many areas where counting strategies are used is in considering the task of reconnecting the wires in a longdistance telephone cable when it has been accidently cut. Suppose a cable contains 120 individual wires that have been cut. (a) How many different ways exist to reconnect the individual wires when the cable is spliced? (b) How much difference would it make in splicing time if the 120 wires in the cable had been grouped into 10 bundles of 12 wires each? (a) There are P(120,120)=120! ways to match up the 120
wires. The value 10!*(12!)^10 is quite a large number. The value 120!, however, is much larger. It has 28 zeros immediately to the left of the decimal point and 199 digits in all. 

15. 
Use algebraic reasoning to show that C(n,r) = C(n1,r) + C(n1,r1). The sequence of steps shown below justifies that we can show that the leftside and rightside expressions in the equation above are equivalent. 