Amy's grandmother gave her 3 identical chocolate chip cookies and 4 identical sugar cookies. In how many different orders can Amy eat the cookies such that either she eats a chocolate chip cookie first, she eats a chocolate chip cookie last, or both?

chocolate chip cookie first: 15 ways

chocolate chip cookie last: 15 ways

chocolate chip cookie first and last: 5 ways

Step-by-step explanation:

There is a total of 7 cookies.

Is she eats a chocolate cookie first, she will have 2 chocolate cookies and 4 sugar cookies (6 in total).

So, to find the number of different orders that Amy can eat the remaining cookies, we just need to calculate a combination of 6 choose 2 (that is, the number of ways we can put the 2 chocolate cookies among the 6 cookies) or a combination of 6 choose 4 (same thing, but for the sugar cookies):

C(6,2) = 6! / (2! * 4!) = 6 * 5 / 2 = 15 ways

Is she eats a chocolate cookie last, she will have 2 chocolate cookies and 4 sugar cookies (6 in total), so the problem is solved again with a combination of 6 choose 2:

C(6,2) = 15 ways

Is she eats a chocolate cookie first and last, she will have just 1 chocolate cookie left and 4 sugar cookies (5 in total), so the problem is solved with a combination of 5 choose 1 or a combination of 5 choose 4:

C(5,1) = 5! / (1! * 4!) = 5 ways

the correct answer is the second option

3< x< 5

step-by-step explanation:

|x - 4| < 1

there are two solutions, one positive and one negative.   remember to flip the inequality when taking the negative solution

x-4 < 1       x-4 > -1

add 4 to each side

x-4+4 < 1+4       x-4+4 > -1+4

x< 5                 x> 3

3< x< 5