How many arrangements of these students on the committee?

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The discussion focuses on calculating the arrangements of students on a committee, emphasizing that members of the same grade must stand together. The proposed solutions involve using factorials, specifically 5! for the groups and 2! for the arrangements within those groups. Additionally, it is noted that fixing one group in the center alters the calculation, leading to the use of 4! for the remaining arrangements. The final arrangement formula suggested is 5! x 2^5, with adjustments for fixed positions. The calculations aim to determine the total number of unique arrangements based on these conditions.
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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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