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The Rotokas of Papua New Guinea have twelve letters in their alphabet. The letters are: A, E, G, I, K, O, P, R, S, T, U, and V. Suppose license plates of five letters utilize only the letters in the Rotoka alphabet. How many license plates of five letters are possible that begin with either G or K, end with T, cannot contain S, and have no letters that repeat?

Oct 16, 2018

#1
+3

Okkaaay. Let us try to solve this one.

First off the problem mentions that the license plate cannot have an "S" in it. Therfore, I am going to take the S away from the list to make our life easier.

A, E, G, I, K, O, P, R, T, U, V...Let's consider this our new list. (We have 11 letters now)

Our license plate is in the form of : $$\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}$$.

-For our first "box", we can only have two choices, since the license plate begins with a G or a K.

-For our second box, we can choose out of $$9$$ letters, since one has already been used above, the letters cannot repeat, and we must reserve "t" to end the plate, as stated in the problem.

-For our third box, we can choose $$8$$ letters, since three have already been used out of the $$11$$ numbers in the new list.

-For our fourth box, following the same pattern as the third, we have $$7$$ choices.

-For our last box, we only have one choice, "t".

So in total, we have $$2 \times 9 \times 8 \times 7 \times 1$$ choices possible.

Calculating that we get 1008 possible license plate combinations, so our answer should be $$\boxed {1008}$$.

Let me know if anything seems off and I will look into it. Oct 16, 2018

#1
+3

Okkaaay. Let us try to solve this one.

First off the problem mentions that the license plate cannot have an "S" in it. Therfore, I am going to take the S away from the list to make our life easier.

A, E, G, I, K, O, P, R, T, U, V...Let's consider this our new list. (We have 11 letters now)

Our license plate is in the form of : $$\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}$$.

-For our first "box", we can only have two choices, since the license plate begins with a G or a K.

-For our second box, we can choose out of $$9$$ letters, since one has already been used above, the letters cannot repeat, and we must reserve "t" to end the plate, as stated in the problem.

-For our third box, we can choose $$8$$ letters, since three have already been used out of the $$11$$ numbers in the new list.

-For our fourth box, following the same pattern as the third, we have $$7$$ choices.

-For our last box, we only have one choice, "t".

So in total, we have $$2 \times 9 \times 8 \times 7 \times 1$$ choices possible.

Calculating that we get 1008 possible license plate combinations, so our answer should be $$\boxed {1008}$$.

Let me know if anything seems off and I will look into it. KnockOut Oct 16, 2018