[A] 21

**24**

[C] 32

[D] 34

[E] 35

[spoiler]OA=C[/spoiler]

Source: Magoosh

[C] 32

[D] 34

[E] 35

[spoiler]OA=C[/spoiler]

Source: Magoosh

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**A**

**B**

**C**

**D**

**E**

Given that the length of each side of a quadrilateral is a distinct integer and that the longest side is not greater than 7, How many different possible combinations of side lengths are there?

[A] 21

** 24**

[C] 32

[D] 34

[E] 35

[spoiler]OA=C[/spoiler]

Source: Magoosh

[A] 21

[C] 32

[D] 34

[E] 35

[spoiler]OA=C[/spoiler]

Source: Magoosh

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- GMAT Instructor
**Posts:**6362**Joined:**25 Apr 2015**Location:**Los Angeles, CA**Thanked**: 43 times**Followed by:**25 members

Recall that in a quadrilateral, the sum of three sides must be greater than the remaining side.

Let’s first determine the number of ways to pick four distinct positive integers not greater than 7; i.e. from among 1, 2, 3, 4, 5, 6, and 7. Since there are 7 integers and we are choosing 4 of them, there are 7C4 = 7!/(4!*3!) = (7x6x5)/(3x2) = 35 ways to make this choice.

Among these 35 possibilities, some of them won’t satisfy the rule we mentioned above. To determine the invalid choices, let’s choose three sides to be as small as possible.

If the three smallest sides are 1, 2 and 3, then we see that the fourth side cannot equal 1 + 2 + 3 = 6 or any length greater than 6. We see that there are 2 invalid choices (which are 1-2-3-6 and 1-2-3-7).

If the three smallest sides are 1, 2 and 4, then we see that the fourth side cannot equal 1 + 2 + 4 = 7 or any length greater than 7. We see that there is 1 invalid choice (which is 1-2-4-7).

If the three smallest sides are not 1-2-3 or 1-2-4, then the sum of the three smallest sides is at least 8 and there are no invalid choices.

Since there are 35 possibilities in total and 3 of them are invalid, the number of possible combinations of side lengths is 35 - 3 = 32.

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