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Permutations and Combinations Generalized Permutations and Combinations Generating Permutations and Combinations

Computer Science 235 Project
Permutations & Combinations Patrick McAtee, Tom Rice, and Jesse Whidden 


Source code: PermComb.java
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1.A clothing company is developing a new shirt design. The shirt will be white and have three stripes down the right side. The colors of the stripes will be green, yellow, and orange. How many ways can these strips be ordered on the shirt?
2.In a race at a track meet, the finals of the 100m dash involve eight runners. However, only six of these runners can earn points for their team. If all finishing orders are possible, how many different ways could the points end up being awarded?
3.A museum has enough open wall space to display either four 4’ x 3’ paintings or two 8’ x 4’ paintings. In their collection, there are 23 paintings of size 4’ x 3’, and 5 paintings of size 8’ x 4’. How many display options are there for the museum?
4.How many permutations can be made from the set {A, E, I, O, U, Y}?
5.Suppose tha a department contains 10 men and 15 women. How Many ways are there to form a committee with 6 members if it must have the same number of men and women?
6.Generating Acronyms: How many threeletter acronyms are possible if every acronym must contain at least one vowel and at least one consonant with no repeating letters?
7.A club has 25 members:
a) How many ways are there to choose four members of the club to serve on an executive committee?
b) How many ways are there to first choose a president, and then a vice president, treasurer and secretary of the club?
8. Which of the following are derangements of 12345?
a) 21435
b) 41352
c) 43152
d) 34521
e) 54321
f) 51432
9.How many derangements of n objects are there for n = ?
a) 4
b) 5
c) 7
d) 9
1.There are six possible ways to order the stripes, they are; (green, yellow, orange); (green, orange, yellow); (orange, green, yellow); (orange, yellow, green); (yellow, orange, green); and (yellow, green, orange).
2.This requires a 6permutation of 8 objects. Use of Theorem 1 gives P(8, 6), which is 8 * 7 * 6 = 336.
3.The answer to this one is the number of possibilities using 4’ x 3’ paintings times the number of possibilities using 8’ x 4’ paintings. That is, P(23, 4) times P(5, 2), which is 212,520 * 20 = 4,250,400. Looks like the curator has some decisions to make!
4.Since all six of the letters will be take each time, the answer is P(6, 6) = 720.
5.Three of the men and three of the women must be selected to keep their numbers even. Since there are 10 men and 15 women, the number of ways of creating this committee is found by C(10, 3) * C(15, 3) = 120 * 455 = 54,600.
6.You are filling three spots with letters, and since the order of these three letters doesn’t matter, they can be chosen in an arbitrary order. For the first spot, a vowel, there are five choices (a, e, i, o, u), which is C(5, 1). There are 21 consonants to pick from for a second spot, so C(21, 1). The final spot leaves us with 24 letters to pick from after taking two out for the other two spots (since repetition is not allowed), which is C(24, 1). This means the answer is C(5, 1) * C(21, 1) * C(24, 1) = 5 * 21 * 24 = 2520.
7.a) Since the order of choosing the members is not relevant (offices are not differentiated), a combination needs to be used. The answer is C(25, 4) = 12,650.
b) In contrast, a permutation is needed here because the order of selection matters. In this situation, a permutation should be used, in this case, P(25, 4) = 303,600.
8.a) no; b) no; c) yes; d) yes; e) no; f) yes
9.a) 9; b) 44; c) 1854; d) 133496
Last updated 12/14/2001