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Chris Lele

SAT Permutations and Combinations Practice Questions

If you’ve read the combination and permutation posts, have a go at the following questions. These combinations/permutations questions are about as hard as any combinations or permutations questions you could see test day. Don’t forget to use the dash method!

1. Sally owns five different blouses. If she cannot wear the same blouse on two consecutive days, how many different ways can she dress on Wednesday and Thursday?

(A)  9

(B)  10

(C)  15

(D) 20

(E)  25

 

2. Quentin has five textbooks on his shelf. If chemistry must always be in the middle, how many ways can he arrange the books?

(A)  24

(B)  48

(C)  72

(D) 112

(E)  120

 

3. Tyler wants to choose an outfit for dinner. If he can choose from 3 shirts, 5 pairs of shoes, and 2 dress pants, how many different outfits can Tyler wear?

(A)  15

(B)  20

(C)  30

(D) 42

(E)  60

 

 

Explanations:

1. The best method is to use the dashes. The first dash can stand for Wednesday and the second for Thursday. For Wednesday Sally can wear any of the five blouses. On Thursday she can only choose from four different blouses. Remember she can’t wear blouses on consecutive days. This gives us: 5 x 4 = 20. Answer (D).

2. The trick here is to imagine that the chemistry book is invisible. After all, it is stuck in one place so it can’t be switched around with the other books. So it’s as though we only have four books. How many ways can we arrange four books? 4! = 24  Answer (A).

3. Tyler is only choosing 1 of each article of clothing. While we could use the combination formula as the first step, whenever you are choosing for only one thing, the number of possibilities is the number you are choosing from. So if he can choose one shirt from 3 shirts, the total number of shirts he can select is 3.   Using the dash method, we can place a number in each dash: 3 x 5 x  2 = 30. Answer (C).

 

 

About Chris Lele

For the last ten years, Chris has been helping students excel on the SAT, ACT and GRE. In this time, he’s coached 5 students to a perfect SAT score. Some of his GRE students have raised their scores by nearly 400 points. He has taken many GMAT students from the doldrums of the 600s to the coveted land of the 700+. Rumor has it he does a secret happy dance when his students get a perfect score.

You can read Chris's awesome blog posts on the Magoosh GRE blog and High School blog!

You can follow him on Twitter and Facebook!


4 Responses to “SAT Permutations and Combinations Practice Questions”

  1. Samuel Phiri says:

    These are great questions I love them, perhaps you can add more. Thank you.

    • Chris Lele Chris Lele says:

      Hi Samuel,

      Sure! In the new few months, I’m sure there will be another lesson on permutations and combinations, one with more questions.

      Stay tuned!

  2. Anes P A says:

    Dear Sir,

    I have a kerala psc 2003 question as “Different Colours 3 skirts, 4 Blouse, 3 Shawl bought by Dhanya. But Green skirt and Green Blouse could not use due to inappropriate look. How many different ways she can wear them?

    (a) 36 (b) 34 (c) 33 (d) 35
    I select 33 ? is it right ?

    Reply me”

    • Magoosh Test Prep Expert Magoosh Test Prep Expert says:

      Hi Anes 🙂

      Yes, you’re correct 😀 Let’s look at how to solve this problem to see why. In this question, we’re dealing with permutations with restrictions. A straight forward way to approach these types of problems is to consider the different scenarios. Here, we have:

      1. Green skirt and not green blouse
      2. Green blouse and not green skirt
      3. Neither green blouse nor green skirt

      We can use permutations to calculate how many different outfits we can have in each of the three situations and then add these three values together to get the total number of possible outfits.

      1. Green skirt and not green blouse
      Number of possible skirts: 1 (green)
      Number of possible blouses: 3 (all minus green)
      Number of possible shawls: 3 (all)
      Number of outfits: 1*3*3 = 9

      2. Green blouse and not green skirt

      Number of possible skirts: 2 (all minus green)
      Number of possible blouses: 1 (green)
      Number of possible shawls: 3 (all)
      Number of outfits: 2*1*3 = 6

      3. Neither green blouse nor green skirt
      Number of possible skirts: 2 (all minus green)
      Number of possible blouses: 3 (all minus green)
      Number of possible shawls: 3 (all)
      Number of outfits: 2*3*3 = 8

      Total number of outfits: 9 + 6 + 8 = 33 outfits

      I hope this helps 🙂

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