How many ways are there for the wedding party to line up in which the bride is not next to the groom

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At a wedding reception, the bride and groom and three attendants will form a receiving line. How many ways can they be arranged in each of following cases? a) Any order will do. b) The bride and groom must be the last two in line. c) The groom must be last in line with the bride next to him.

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Hi I'm David. Your question. Now let me bring up your question here in the question here and the wedding reception. The bride and the groom and the three attendants will form a receiving line. And when you find how many ways can there be arranged in each of the following cases. Now let me go on the Line. 2 3456. And we have totally will be only five of them because we have the bride and the groom and the three attendants. So we have totally five people Now in the question once you have in any other, so because we don't care any other. So therefore this one will be the first chance here. It will be the first uh prohibition. It will be five options will be in the five and then we move on to the next one. We have only four options left. And then for the next one we have the three options. After that we have the two options and the last one will be the last person. And therefore if we do this one we should get the answer. It will equal to you have five times four times three times two temp's one. Mhm. And then we got No 120. And then for the question be we want to have the bride and the groom fold must be on the last two lines. And again we have 1, 2, 3, 4 and five. So the last one. So it means that these two the last two positions here, we've been reserved from the bride and the groom. So this would have been the bright I am groom. And it means that the first three positions would be any other told him. The three charges found the first one, the second the second one have the two chances. This is gonna be the last one. So the brand, the broom, there will be either they can interchange the position so therefore the answer, it would equal to the three times two times one and because they can entertain the two others and therefore we go time still here And then we get the answer equal to six times two is equal to the 12. And for the question see we want to have the crew must be last in line and then right next to him. So we have 1234 and five. So in this case here we want this will be the groom and this will be the brand. So we have only the choices for the first three of them. So this one will be the three choices. This one you have the two choices and this one will have the last one. So then found. The answer will be the three times two times one and then go to the sixth. And that's gonna be the answer for this question here. Right

10.6.3: Choosing a lineup for a traveling basketball team.

There are 20 members of a basketball team.

(a) The coach must select 12 players to travel to an away game. How many ways are there to select the players who will travel?

(b) From the 12 players who will travel, the coach must select her starting line-up. She will select a player for each of the five positions: center, right forward, left forward, right guard, left guard. How many ways are there for her to select the starting line-up?

(c) From the 12 players who will travel, the coach must select her starting line-up. She will select a player for each of the five positions: center, right forward, left forward, right guard, left guard. However, there are only three of the 12 players who can play center. Otherwise, there are no restrictions. How many ways are there for her to select the starting line-up?

(a) C(52, 5)

(b) C(13, 2).C(39, 3)

(c) C(26, 5)

(d) C(13, 1).C(48, 1)

(e) C(13, 1).C(12, 1).C(4,3).C(4,2)

(f) C(13,5).(4^5)

I'm working on some permutation problems for my discrete math course and finding that I'm not fully understanding the material. I've watched a couple YouTube videos and read a couple different math books, and it's still not fully solidifying. I was hoping I could post my problems and that someone could help me think through the problems so that I may better understand the material. Here are the current problems and my current answers:

Ten members of a wedding party are lining up in a row for a photograph.

Problem #1:
How many ways are there to line up the ten people?

Answer for Proble #1:
$10! = 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 3,628,800$

Problem #2:
How many ways are there to line up the ten people if the groom must be to the immediate left of the bride in the photo?

Answer for Question #2:
$9! = 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 362,880$

Problem #3:
How many ways are there to line up the ten people if the bride must be next to the maid of honor and the groom must be next to the best man?

Answer for Question #3:
We can group together the bride and maid-of-honor into one unit with 2 order possibilities: {(bride,maid-of-honor), (maid-of-honor, bride)}, and we can also group together the groom and best main into one unit with 2 order possibilities: {(groom, best man), (best man, groom)}.

Essentially, since these units are grouped together and must stay with one another when ordering, instead of 10 spots in the arrangement, we can think of there actually being 8 spots. For each permutation, there will be 3 extra permutations since the members of each group can swap places. For instance, labeling each member as a number and placing the grouped units at the end of the lineup we get 4 different permutations:

Ultimately, we have the following total permutations:
$8! \cdot 4 = \left (8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \right ) \cdot 4 = 161,280$

Any help is greatly appreciated. Thank you all so much! I hope you all are doing well and staying safe.

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