The Student Room Group

S1 Permutations Questions

Got my exam today and I still can't grasp permutations properly. Can anyone help me with these questions?

A team consisting of 7 players is to line up for a photograph in a single line. The photographer insists that the two tallest players cannot stand together. How many possible line-ups are there?

Also, I tried to work this one out.. But because it says "At least" it's kind of thrown me, as the teams could have 4, 5 or 6 female members.

A class consists of 12 female and 8 male students. 6 students are to be selected to take part in a quiz. How many different teams would be possible if the team had to have at least 4 female members?

12C4 * 8C2 would be the way to go if it was 4 females, but the at least bit confuses me. Help?
Reply 1
Coda
Got my exam today and I still can't grasp permutations properly. Can anyone help me with these questions?

A team consisting of 7 players is to line up for a photograph in a single line. The photographer insists that the two tallest players cannot stand together. How many possible line-ups are there?

Also, I tried to work this one out.. But because it says "At least" it's kind of thrown me, as the teams could have 4, 5 or 6 female members.

A class consists of 12 female and 8 male students. 6 students are to be selected to take part in a quiz. How many different teams would be possible if the team had to have at least 4 female members?

12C4 * 8C2 would be the way to go if it was 4 females, but the at least bit confuses me. Help?



Errm let me try (I have the exam today too) - can someone please correct me it im wrong (just quote me)

If two of the tallest player (X &Y) cannot stand together then it is X _ _ _ _ _ Y or Y _ _ _ _ _ _ X.
So the answr is 2! (as there are two ways for X & Y to stand) x 5! (as there are 5x4x3x2x1 ways in which the others can stand). This = 240.

At least means from 4 females to the max so you would work out 4, 5 & 6 females (there is no restriction on the max number.

So that would be (12C4 * 8C2) + (12C5 * 8C1) + (12C6 * 8C0)

EDIT: Just quoted you :smile:
Reply 2
for the second question you have given all the right cases.... so you know that you have to consider the case for 4, 5 or 6 females ....

you have done the case for 4 females correctly
now do the same but for 5 females and 1 male
then do the same for 6 females and 0 males...

now add all 3
Reply 3
Coda
Got my exam today and I still can't grasp permutations properly. Can anyone help me with these questions?

A team consisting of 7 players is to line up for a photograph in a single line. The photographer insists that the two tallest players cannot stand together. How many possible line-ups are there?

Also, I tried to work this one out.. But because it says "At least" it's kind of thrown me, as the teams could have 4, 5 or 6 female members.

A class consists of 12 female and 8 male students. 6 students are to be selected to take part in a quiz. How many different teams would be possible if the team had to have at least 4 female members?

12C4 * 8C2 would be the way to go if it was 4 females, but the at least bit confuses me. Help?


For the first one consider how many ways there are of standing the two tallest players together, so we can consider them as one person.

__ _ _ _ _ _ where __ is the two tallest players.

You can have 6 positions for this. In each position there are 2! x 5! arrangements.

So 6 x 2! x 5! = 1440.

The total number of arrangements is 7! = 5040.

Hence 5040-1440=3600 ways where they aren't standing together.

For the second one you have to chose 4 females from 12 (doesn't matter about the order, they're still part of the team) so it's 12C4. Then you have to pick 2 males from 8, 8C2. You can also have 12C5 and 8C1 or 12C6.

So the answer is 12C4 x 8C2 + 12C5 x 8C1 + 12C6 = 21,120

I hope these are right, not sure about first one but heck! Do you have the answers?
Reply 4
dime_piece
Errm let me try (I have the exam today too) - can someone please correct me it im wrong (just quote me)

If two of the tallest player (X &Y) cannot stand together then it is X _ _ _ _ _ Y or Y _ _ _ _ _ _ X.
So the answr is 2! (as there are two ways for X & Y to stand) x 5! (as there are 5x4x3x2x1 ways in which the others can stand). This = 240.

At least means from 4 females to the max so you would work out 4, 5 & 6 females (there is no restriction on the max number.

So that would be (12C4 * 8C2) + (12C5 * 8C1) + (12C6 * 8C0)

EDIT: Just quoted you :smile:


For the first one, if there were no restrictions there would be 7! ways to line up the players, so that's 5040. Because there are restrictions, you can treat the two players standing together as "one player" so it's 6! = 720. Because those 2 players can swap between eachother it's 6! x 2 = 1440. So you do 5040 - 1440 = 3600

The other one is completely understandable now :biggrin:
Reply 5
CHEM1STRY
For the first one consider how many ways there are of standing the two tallest players together, so we can consider them as one person.

__ _ _ _ _ _ where __ is the two tallest players.

You can have 6 positions for this. In each position there are 2! x 5! arrangements.

So 6 x 2! x 5! = 1440.

The total number of arrangements is 7! = 5040.

Hence 5040-1440=3600 ways where they aren't standing together.

For the second one you have to chose 4 females from 12 (doesn't matter about the order, they're still part of the team) so it's 12C4. Then you have to pick 2 males from 8, 8C2. You can also have 12C5 and 8C1 or 12C6.

So the answer is 12C4 x 8C2 + 12C5 x 8C1 + 12C6 = 21,120

I hope these are right, not sure about first one but heck! Do you have the answers?


You were correct. Darn VLE quiz :biggrin:
Reply 6
Coda
You were correct. Darn VLE quiz :biggrin:


Lol I don't often get them right :frown: Everyone finds them hard!!

Anyway hope I helped :biggrin:
Reply 7
Coda
For the first one, if there were no restrictions there would be 7! ways to line up the players, so that's 5040. Because there are restrictions, you can treat the two players standing together as "one player" so it's 6! = 720. Because those 2 players can swap between eachother it's 6! x 2 = 1440. So you do 5040 - 1440 = 3600

The other one is completely understandable now :biggrin:


Oh i get it know! Thanks :smile:

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