Question

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(B) 24(C) 32(D) 34(E) 35

Answer 1

Answer:

(E) 35

Step-by-step explanation:

The possible lengths for a side are: {1, 2, 3, 4, 5, 6, 7}

Therefore 7 possible lengths.

The quadrilateral has 4 sides.

It is clearly stated that the sides lengths should be distinct, so it indicates we are dealing with either permutations or combinations.

The order here does not matter. For example the quadrilateral with sides {1, 2, 3, 4} is the same quadrilateral as if it was {4, 3, 2, 1}

Therefore we are dealing with combinations.

The formula is:

$\displaystyle\frac{n!}{(n-r)!r!}$

Here n is 7 and r is 4, since we have a group of 7 items to pick (lengths) and we will form groups of 4 items (quadrilaterals)

Therefore the formula becomes:

$\displaystyle\frac{7!}{(7-4)!4!}=\frac{7!}{3!4!}$

We re-write 7! as 7(6)(5)4! to be able to cancel the 4!:

$\displaystyle\frac{7(6)(5)\not{4!}}{3!\not{4!}}=\frac{7(6)5}{3!}$

And notice 3! is 3(2)(1)=6 so we finally get:

$\displaystyle\frac{7(\not{6})(5)}{\not{6}}=7(5)=35$

Therefore there are 35 possible combinations of side lengths