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# Arrangements watch

1. (Original post by TeeEm)
firstly treat the 1st years as one lump (inseparable) and the 3 2nd years separately.

work out this perm

then how many perms can you get from the lump of 5...

etc
First years as inseparable? Hm won't that be 3! = 6 ?
2. (Original post by Affection)
First years as inseparable? Hm won't that be 3! = 6 ?
5 women 5 men

how many ways 5 women sit next to each other?

5 men + 1 LUMP = 6

6! ways = 720 ways

but in each of these 720 ways the LUMP can appear 5! different ways = 120 ways

therefore

120 x 720 ways the 5 women sit next to each other
3. (Original post by TeeEm)
5 women 5 men

how many ways 5 women sit next to each other?

5 men + 1 LUMP = 6

6! ways = 720 ways

but in each of these 720 ways the LUMP can appear 5! different ways = 120 ways

therefore

120 x 720 ways the 5 women sit next to each other
Oh... um for part a) its: 4! = 24 x 5! = 120 ?
I think?
Btw i'm sorry if i'm asking you too many questions :s I tend to understand things slowly
4. (Original post by Affection)
Oh... um for part a) its: 4! = 24 x 5! = 120 ?
I think?
Btw i'm sorry if i'm asking you too many questions :s I tend to understand things slowly
i see ...

24 x 5! is not 120

(a bit larger than 120)
5. (Original post by TeeEm)
i see ...

24 x 5! is not 120

(a bit larger than 120)
oops my mistake its supposed to be 2880 >_<
6. (Original post by Affection)
oops my mistake its supposed to be 2880 >_<
something like that
7. (Original post by TeeEm)
something like that
So would part B) be 2! ??
8. (Original post by Affection)
So would part B) be 2! ??
now 2 lumps and each lump has its rearrangements, extend it ...
9. (Original post by TeeEm)
now 2 lumps and each lump has its rearrangements, extend it ...
extend it???
10. (Original post by Affection)
extend it???
it is practically an extension of the previous part

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