You are going to make “words” using the letters in the word WHISKAS. a) How many seven-letter words can you make? b) How many seven-letter words can you make if the two S’s must be together? c) How many seven-letter words can you make if the words must begin with A and end with H?

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PART A

When we say words, we don't really mean words. We just mean how many different ways we can arrange the letters. We could make a word like Hiskasw and that would work.

We have six different letters: w, h, i, s, k, and a. We have two s's and this will be very important when finding these permutations.

The first thing we do is take the number of letters and find its factorial. In this case, it is seven, so we have 7×6×5×4×3×2×1=5040.

But, this is not how many combinations there are, because we have two s's, and since they are the same many of our combinations are actually identical, but the s's are just switched. So, we take the number of s's (2) and we factorial it, which just still equals two, and then we divide our first factorial by that.

5040/2=2520

There are 2520 ways you can arrange the word WHISKAS.

PART B

Let's think of the seven letter slots in the word. It doesn't matter where you place one of the S's, but a slot next to the first S you place only has one choice of letter, which is another S to make it adjacent. This means that one  specific slot is required to have an S. If we start off filling in our required one with an S, we have six letters left to fill in our other slots, which will give us a result of 6!, which is 720 seven-letter words.

PART C

Now, we have to have A be the first term and H be the last. If we think of the seven letter slots, we have only one choice on the first one, which is A, and only one choice on the last one, which is H. This leaves us with 5! or 120 possibilities, but we also have to divide by two because we have two S's, so there are 60 seven-letter words you can make if the words must begin with A and end with H.

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answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204