Perms and Combs (1 Viewer)

[thoth1]

can sumone confirm my answer to this question:

A family of six are sitting around a round table in how many ways can they be arranged so that mum and dad are sitting together and youngest daughter does not sit next to dad.

I got 2!x2!x3P2 = 24 ways

can sumone confirm if this is right bcuz this was in a Q7 of a 4U trial. just want to make sure there arent any triks.

 

[tambam]

can sumone confirm my answer to this question:

A family of six are sitting around a round table in how many ways can they be arranged so that mum and dad are sitting together and youngest daughter does not sit next to dad.

I got 2!x2!x3P2 = 24 ways

can sumone confirm if this is right bcuz this was in a Q7 of a 4U trial. just want to make sure there arent any triks.

Okay fine, because you asked nicely, i will attempt to do your question, though i can't guarantee that its correct, and anyone feel free to correct me.

This is what i did:
What are the possible permutations where the mother(M) and father(F) are seated together, which would be 4!x2= 48

Then you need, in how many of those permutations is the youngest daughter NOT sitting next to father?
You can find this out by doing, P(mother&father together)-P(father sitting next to daughter)
So the number of permutations where the father and daughter(D) are together is 3!x2
(Group MDF together, circle perm becomes 3!, then within the group of 3 you can have MFD or DFM where the father and daughter are together, and father and mother are still together)

Giving an answer of 4!x2-3!x2= 36

But yeah, like i said, i'm not excellent at maths. So i could be incorrect,but i don't understand the method you used.

 

[pwoh]

Got the same answer as tambam (different method though)

[Arrange mother & father] x [Place younger daughter] x [Place other 3 children]
2x3x3! = 36