Right and Wrong Permutations and Combinations Methods And Pointers That Will Make You Ace These On Your CSET Exam

4. Seven cards are chosen from a standard 52-card deck. How many of these will have four diamonds and three hearts?

A. 29,446,560

B. 204,490

C. 1,001

D. None of the above

5. Determine whether the following situation would require calculating a permutation or a combination: Selecting three students to attend a conference in California.

6. Determine whether the following situation would require calculating a permutation or a combination: Selecting a stage performer and a sound man for a school play.

7. Determine whether the following situation would require calculating a permutation or a combination: Assigning students to their seats on the first day of school.

Screen shot of CSET Math exercise on permutations

Answer Key

1. A (This problem involves a combination of 20 people taken 4 at a time. It is not a permutation because the order in which the people are selected does not matter. To get the answer, you must find C (20,4) = (20*19*18*17)/4! = 4,845)

2. C (This problem involves a permutation of 20 people taken 4 at a time. It is not a combination because the order in which the people are selected matters. To obtain the answer, you need P (20,4) = 20*19*18*17 = 116,280)

3. A (The solution involves the multiplication principle and permutations. First three letters must be chosen from 26, and since order is important, this is P (26,3). Then two digits must be chosen from 10, and this is P (10,2). By the multiplication principle, the number of possible serial numbers is P (26,3) * P (10,2) = 26*25*24*10*9 = 1,404,000.

4. B (This problem involves choosing 4 diamonds from 13 followed by choosing 3 hearts from 13. The order in which the cards are selected does not matter. The possible number of hands with these characteristics is C (13,4) * C (13,3) = 204,490.

5. combination, 6. permutation, 7. permutation