perms and combs (1 Viewer)

[eternallyboreduser]

Why are the answers 40320, 20160 and 17280?

I did this:
a)8c4×4!×4c4×4!/2!=20160
b)1×7c3×4c4×4!×4!/2!=10080
c)1×7c2×4c4×4!×4!/2!=6048

8 ppl are to form 2 queues of 4, in how many ways can this be done if:
a)no restrictions
b)jim only stands in left queue
c)sean and liam must stand in the same queue

 

[cossine]

Why are the answers 40320, 20160 and 17280?

I did this:
a)8c4×4!×4c4×4!/2!=20160
b)1×7c3×4c4×4!×4!/2!=10080
c)1×7c2×4c4×4!×4!/2!=6048

8 ppl are to form 2 queues of 4, in how many ways can this be done if:
a)no restrictions
b)jim only stands in left queue
c)sean and liam must stand in the same queue

a) is very easy. Notice 8! = 40320.

 

[liamkk112]

Why are the answers 40320, 20160 and 17280?

I did this:
a)8c4×4!×4c4×4!/2!=20160
b)1×7c3×4c4×4!×4!/2!=10080
c)1×7c2×4c4×4!×4!/2!=6048

8 ppl are to form 2 queues of 4, in how many ways can this be done if:
a)no restrictions
b)jim only stands in left queue
c)sean and liam must stand in the same queue

for a and b, no need to divide by 2!, otherwise your reasoning seems correct judging from the numbers you’ve used.

for c) you have two cases; they’re both in the first queue and they’re both in the second queue. now there are 2 ways to choose which queue they’re both in, and 6C2 ways to pick two more people to go into the queue with them both, and 4! ways to organise that queue. there are then 4C4 x4! ways to do the other queue, which gives a total of 2x6C2x4!x4C4x4!= 17280 ways