SOLUTION: Could someone please help me with this problem? I think I may be on the right track, but am not really sure! "From a squad of 12 cheerleaders, 10 will assemble themselves into a 4

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Question 1029837: Could someone please help me with this problem? I think I may be on the right track, but am not really sure!
"From a squad of 12 cheerleaders, 10 will assemble themselves into a 4-level pyramid.
a) how many different combinations of cheerleaders can be used to build the pyramid? **I think it's 12/10 or 12 C 10 (sub 12, C, sub 10/don't know how to get it on here correctly, sorry!) which equals 66, but I don't understand how to work it out.
b) the coach decides that Alexis must be at the top of the pyramid, and that Rachel, Jen, Britt, and Nicole will form the base. How many different combinations of 10 cheerleaders can now be chosen to form the pyramid?
**I think it's 7/5 and answer is 21, but again I don't know for sure or how to actually arrive at that answer
c) Suppose alexis has an injury and can't participate. the coach replaces Alexis with Rebecca. Now, how many different combinations of 10 cheerleaders can the coach select?
*I think it's 6/5 which equals 6, same on this one, not sure how to work out. Or, if my answers are even correct??
Any help would be appreciated! Thanks!
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
There are originally 12 cheerleaders that can participate.
Of those 12 , 10 will form a pyramid.
The other 12-10=2 will just watch.
As the problem's wording suggests, it is a question of combinations (where order/position in the pyramid does not matter).
It is not a question of permutations.

a) The number of different combinations (subsets) of 10 cheerleaders
that can be made from that set of 12 available cheerleaders can be calculated by a formula or just reasoned through.
The written and applied formula may be required by the teacher.
Reasoning is all that is needed to get to the solution and/or the formula.
The formula for the number of combinations (subsets) of r objects is
%28matrix%282%2C1%2Cn%2Cr%29%29=n%21%2F%28%28n-r%29%21r%21%29 .

In this case,
.
You could look at it from the other side.
The number of different possible combinations of 10 cheerleaders to be used to make the pyramid is the same as
the number of different possible combinations of 2 disappointed cheerleaders can be chosen to not participate in the pyramid.
Reasoning and calculating the answer that way is easier.
The coach could look at the cheerleaders and decide to leave out the 2 of them who have missed the most practices.
There would be 12 ways to make his first choice of cheerleader to exclude,
and 11 ways to make the second choice.
So the coach's thinking could be made in 12%2A11 ways when order of choices matter.
But since the same set of 2 excluded cheerleaders could be picked 2 different ways,
the number of different possible sets of 2 excluded cheerleaders is
12%2A11%2F2=66 .
Applying the formula, we would calculate it as
%28matrix%282%2C1%2C12%2C2%29%29=12%21%2F%282%21%2812-2%29%21%29=12%21%2F%282%2110%21%29=66

b) Once the coach decide on the 5 cheerleaders that will be the top and the base of the pyramid,
it is a question of how many different sets of other cheerleaders can the coach choose to form the other two layers to complete the pyramid.
There are still 12-5=7 cheerleaders to choose from for the other 10-5=5 pyramid positions.
The number of possible choices is
%28matrix%282%2C1%2C7%2C5%29%29=7%21%2F%28%287-5%29%215%21%29=7%21%2F%282%215%21%29=21 .
You can also reason there are 7%2A6%2F2=21 ways to pick the two cheerleaders who will not participate.

c)With Alexis injured, there are only 12-1=11 available cheerleaders.
Other than that, it is like the situation in part b:
after the coach picks 5 cheerleaders for specific positions in the 10 person pyramid,
there are 10-5=5 more cheerleaders to be chosen
out of the 11-5=6 remaining available cheerleaders.
%28matrix%282%2C1%2C6%2C5%29%29=6%21%2F%28%286-5%29%215%21%29=6%21%2F%281%215%21%29=61 .
You can also reason there are 6 ways to pick the 1 of the 6 cheerleaders who will not participate.

THE REASONING FOR THE FORMULA:
The reasoning is that when you make the list of the r objects,
there are n possibilities for the first object to be listed,
n-1 possibilities for the second one, and so on,
so the list is being built n%2A%28n-1%29%2A%22...%22 ways.
The number of ways is a product of consecutive factors counting down from n .
After the last item is chosen, there are n-r objects not chosen,
so the final product does not include %28n-r%29%2A%28n-r-1%29%2A%22...%22%2A3%2A2%2A1=%28n-r%29%21 ,
but all the other factors in n%21=n%2A%28n-1%29%2A%22...%22%2A3%2A2%2A1 are included.
The final product is
n-%28n-1%29%2A%22...%22%2A%28n-r%2B1%29= , and that is the number of ways the list could be built.
However, a list of r objects can be written r%21=r%2A%28r-1%29%2A%22...%22%2A3%2A2%2A1 different ways,
because there are r items that could be first, and for each case, there would be r-1 items that could be second, and so on.
Since the n%21%2F%28n-r%29%21 lists include r%21 repeats of each combination/subset of r objects,
the number of combinations of r objects that can be made from a set of n objects is