How many arrangements of these students on the committee?

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SUMMARY

The discussion focuses on calculating the number of arrangements for students on a committee, specifically addressing the requirement that members of the same grade must stand together. The solution involves using factorial calculations, specifically 5! for the groups and 2! for the arrangements within each group. Additionally, when one group (Grade 12) is fixed in the center, the calculation adjusts to 4! x 2^5, accounting for the arrangements of the remaining groups. The final arrangement formula is confirmed as 4! x 2^5.

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Homework Statement
Arrange members of committee
Relevant Equations
fundamental counting principle
1651331490818.png

11.1 10 x 9

11.2.1 10!

11.2.2 I'm not sure.

Attempt at solution:

a) members of same grade must stand together => 5! x 2!
b) Grade 12 in the centre means 1 of 5 groups is fixed but there are 2 ways the centre group can stand => 4! x 2! x2!
 
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neilparker62 said:
ttempt at solution:

a) members of same grade must stand together => 5! x 2!
b) Grade 12 in the centre means 1 of 5 groups is fixed but there are 2 ways the centre group can stand => 4! x 2! x2!
Each of the others grades can stand in two ways as well.
 
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PeroK said:
Each of the others grades can stand in two ways as well.
So with 5 groups of two:

5! x 2^5

and with one of those in a fixed position:

4! x 2^5 ?
 
neilparker62 said:
So with 5 groups of two:

5! x 2^5

and with one of those in a fixed position:

4! x 2^5 ?
That looks right.
 
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