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Worksheet on Combinations and Permutations - Solutions

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Permutations and combinations are easy to get confused. They are both ways to count r things taken from a set of n distinct objects, but the way that we count is different.

The key thing to remember is that permutations deal with situations when the order that we choose the objects or arrange them is important. In combinations we are not concerned with what order we selected our objects. We only need this concept, and the formulas for combinations and permutations to solve problems dealing with this topic.

Here are some practice problems with solutions to help you straighten out the ideas of permutations and combinations. A version without answers is here.

  1. Calculate P( 5, 2 ).

  2. SOLUTION: Here we must simply use the formula for permutations. Be sure to remember how to calculate factorials: P( 5, 2 ) = 5!/3! = 20

  3. Calculate C( 5, 2 ).

  4. SOLUTION: Here we must simply use the formula for combinations C( 5, 2 ) = 5!/(2!3!) = 20/2 = 10.

  5. Calculate P( 6, 6 ).

  6. SOLUTION: The formula for permutations, along with the fact that 0! = 1, gives us P( 6, 6 ) = 6!/0! = 720.

  7. Calculate C( 6, 6 ).

  8. SOLUTION: The formula for combinations, along with the fact that 0! = 1, gives us C( 6, 6 ) = 6!/(6!0!) = 1.

  9. Calculate P( 100, 97 ).

  10. SOLUTION: The formula for permutations gives us P( 100, 97 ) = 100!/97!. This expression will be easier to evaluate if we rewrite it as follows: 100!/97! = (100 x 99 x 98 x 97!)/97! = 100 x 99 x 98 = 970200.

  11. Calculate C( 100, 97 ).

  12. SOLUTION: The formula for permutations gives us C( 100, 97 ) = 100!/(97!3!). This expression will be easier to evaluate if we rewrite it as follows: 100!/97! = (100 x 99 x 98 x 97!)/(97!3!) = (100 x 99 x 98)/6 = 970200/6 = 161700.

  13. It’s election time at a high school that has a total of 50 students in the junior class. How many ways can a class president, class vice president, class treasurer,and class secretary be chosen if each student may only hold one office?

  14. SOLUTION: Here the order that we choose the students is important. The same group of four people could be given different offices and it would count as a different result. Since order matters, we are really trying to determine the number of permutations of 50 taken four at a time. The formula for permutations gives us P( 50, 4 ) = 50!/46! = 50 x 49 x 48 x 47 = 5527200.

  15. The same class of 50 students wants to form a prom committee. How many ways can a four person prom committee be selected from the junior class?

  16. SOLUTION: This is similar to the last example in that the numbers are the same, we are choosing four people out of 50. However this time the order is unimportant. If a group of people is on the committee, then it does not matter what order they were selected. Since order does not matter, we are trying to determine the number of combinations of 50 taken four at a time. The formula for combinations gives us C( 50, 4 ) = 50!/(4!46!) = (50 x 49 x 48 x 47)/(4 x 3 x 2 x 1) = 230300.

  17. If we want to form a group of five students and we have 20 to choose from, how many ways is this possible?

  18. SOLUTION: In forming a group we don’t need to worry about the order. So this is a combination and we need to calculate C( 20, 5 ) = 20!(5!15!) = (20 x 19 x 18 x 17 x 16)/ (5 x 4 x 3 x 2 x 1) = 15504.

  19. How many ways can we arrange four letters from the word “computer” if repetitions are not allowed, and different orders of the same letters count as different arrangements?

  20. SOLUTION: Since we’re keeping track of the order of the letters, this is a permutation. We are selecting four letters from a total of eight, and must calculate P( 8, 4 ) = 8!/4! = 8 x 7 x 6 x 5 = 1680

  21. How many ways can we arrange four letters from the word “computer” if repetitions are not allowed, and different orders of the same letters count as the same arrangement?

  22. SOLUTION: Order is not important, so this is a combination. We are selecting four letters from a total of eight, and calculate C( 8, 4 ) = 8!/(4!4!) = (8 x 7 x 6 x 5)/(4 x 3 x 2 x 1) = 70

  23. How many different four digit numbers are possible if we can choose any digits from 0 to 9 and all of the digits must be different?

  24. SOLUTION: Here the order is important, so this is a permutations. We must calculate P( 10 , 4) = 10!/6! = 10 x 9 x 8 x 7 = 5040.

  25. If we are given a box containing seven books, how many ways can we arrange three of them on a shelf?

  26. SOLUTION: Here we care about how the books are arranged, and so the order is important. This means that we are dealing with a permutation. We must calculate P( 7 , 3) = 7!/4! = 7 x 6 x 5 = 210.

  27. If we are given a box containing seven books, how many ways can we choose collections of three of them from the box?

  28. SOLUTION: Here we care about choosing groups of three books, and so the order is not relevant to the problem. This means that we are dealing with a combination. We must calculate C( 7 , 3) = 7!/(4!3!) = 7 x 6 x 5 = 35.

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