Question 1029837
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
{{{(matrix(2,1,n,r))=n!/((n-r)!r!)}}} .


In this case,
{{{(matrix(2,1,12,10))=12!/((12-10)!10!)=12!/(2!10!)=12*11*10*9*8*7*6*5*4*3*2*1/((2*1)(10*9*8*7*6*5*4*3*2*1))=12*11/2=66}}} .
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*11}}} 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*11/2=66}}} .
Applying the formula, we would calculate it as
{{{(matrix(2,1,12,2))=12!/(2!(12-2)!)=12!/(2!10!)=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
{{{(matrix(2,1,7,5))=7!/((7-5)!5!)=7!/(2!5!)=21}}} .
You can also reason there are {{{7*6/2=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.
{{{(matrix(2,1,6,5))=6!/((6-5)!5!)=6!/(1!5!)=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*(n-1)*"..."}}} 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 {{{(n-r)*(n-r-1)*"..."*3*2*1=(n-r)!}}} ,
but all the other factors in {{{n!=n*(n-1)*"..."*3*2*1}}} are included.
The final product is
{{{n-(n-1)*"..."*(n-r+1)}}}={{{n*(n-1)*"..."*(n-r+1)*(n-r)*(n-r-1)*"..."*3*2*1/((n-r)*(n-r-1)*"..."*3*2*1)=n!/(n-r)!}}} , and that is the number of ways the list could be built.
However, a list of {{{r}}} objects can be written {{{r!=r*(r-1)*"..."*3*2*1}}} 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!/(n-r)!}}} lists include {{{r!}}} 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
{{{(matrix(2,1,n,r))=((n!/(n-r)!))/r!=n!/(n-r)!r!)}}}