did i solve these counting problems correctly?

Question

nine people (Ann, Ben, Cal, Dot, Ed, Fran, Gail, Hal, and Ida) are in a room. Five of them stand in a row for a picture.

a) In how many ways can this be done if both Ed and Gail are in the picture?

P(5,2)= 5×4 = 20

b) In how many ways can this be done if neither Ed nor Fran are in the picture?

P(5,3)= 5×4×3= 60

c) In how many ways can this be done if Dot is on the left end and Ed is on the right end? P(5,3)= 5×4×3= 60

d) In how many ways can this be done if Hal or Ida (but not both) are in the picture? P(5,4) + P(5,4)= 120+120= 240

e) In how many ways can this be done if Ed and Gail are in the picture, standing next to each other? P(5,4)= 120

Thank you

Answer 1

I post the answer with explanation for each section below. Hope it will be clear enough.

a) In how many ways can this be done if both Ed and Gail are in the picture?

Given that Ed and Gail are already in picture, you need 3 more from the rest of the people in ${7 \choose 3}$ =35 ways. Also, 5 people can stand in a picture in $5!$ ways. So the answer is $35*120=4200$.

b) In how many ways can this be done if neither Ed nor Fran are in the picture?

Since, you need to choose 5 from 7 people, you can do it in ${7 \choose 5} = 42$ ways. Also, since ordering matters, you need to multiply this value by $5!=120$. Answer: $42*120=5040$.

c) In how many ways can this be done if Dot is on the left end and Ed is on the right end?

Now you need to feel 3 spots in the middle of picture from 7 people that are not already in picture. That is ${7 \choose 3}=35$. While seats on the left and right are reserved, you can still arrange the people in the middle in $3!=6$ ways. Hence, asnwer: $35*6=210$.

d) In how many ways can this be done if Hal or Ida (but not both) are in the picture?

We can calculate in different ways, but the way I do is to calculate the case Hal is in the picture but not Ida, and Ida is in the picture but not Hal cases separately, and add them up. Hal is in picture, but not Ida: ${7 \choose 4}=35$ ways of choosing people, and $5!$ ordering. Answer: $35*120=4200$ Ida is in picture, but not Hal: Exactly the same as above with the names flipped. Hence, answer: 4200. Overall: 8400

In how many ways can this be done if Ed and Gail are in the picture, standing next to each other? 

Similarly, here you need to choose 3 people from the rest (${7 \choose 3}=35$. But this time, we will treat Ed and Gail as a single body, and the number of different orderings for 4 people is $4!$. However, since, Ed and Gail can exchange their seats with each other, we still need to multiple overall result to 2. Hence, answer: $35*24*2=960$.