To solve this problem, we must understand the concept of combination/permutations. A combination is used when the order doesn't matter while a permutation is used when order matters. In this problem, the two class representatives are randomly chosen, therefore it doesn't matter what order the representative is chosen in, the end result is the same.  The general formula for combinations is \(\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}\), where \(\displaystyle n\) is the number of things you have and \(\displaystyle k\) is the things you want to combine.

Plugging in choosing 2 people from a group of 20, we find

\(\displaystyle C(20,2)=\frac{20!}{(20-2)!2!}=190\).  Therefore there are a \(\displaystyle 190\) different ways to choose the \(\displaystyle 2\) class representatives.

The number of potential combinations for \(\displaystyle k\) selections made from \(\displaystyle n\) possible options is

\(\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}\)

Quantity A:

\(\displaystyle C(8,4)=\frac{8!}{(8-4)!4!}=70\)

Quantity B:

\(\displaystyle C(8,5)=\frac{8!}{(8-5)!5!}=56\)

Quantity A is greater.

For \(\displaystyle k\) choices made from \(\displaystyle n\) possible options, the number of potential combinations (order does not matter) is

\(\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}\)

And the number of potential permutations (order matters) is

\(\displaystyle P(n,k)=\frac{n!}{(n-k)!}\)

Quantity A:

\(\displaystyle C(10,4)=\frac{10!}{(10-4)!4!}=210\)

Quantity B:

\(\displaystyle P(10,2)=\frac{10!}{(10-2)!}=90\)

Quantity A is greater.

When dealing with combinations, the number of possible combinations when selecting \(\displaystyle k\) choices out of \(\displaystyle n\) options is:

\(\displaystyle C(n,k)=\frac{n!}{(n-k)! k!}\)

For Quantity A, the number of combinations is:

\(\displaystyle C(10,3)=\frac{10!}{(10-3)!3!}\)

\(\displaystyle C(10,3)=120\)

For Quantity B, the number of combinations is:

\(\displaystyle C(10,5)=\frac{10!}{(10-5)!5!}\)

\(\displaystyle C(10,5)=252\)

Quantity B is greater.

Since in this problem we're dealing with combinations, the order of selection does not matter.

With \(\displaystyle k\) selections made from \(\displaystyle n!\) potential options, the total number of possible combinations is

\(\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}\)

Quantity A:

\(\displaystyle C(10,2)=\frac{10!}{(10-2)!2!}=45\)

Quantity B:

\(\displaystyle C(20,4)=\frac{20!}{(20-4)!4!}=4845\)

Quantity B is larger.

Since we're dealing with combinations in this problem, the order of selection does not matter.

With \(\displaystyle k\) selections made from \(\displaystyle n!\) potential options, the total number of possible combinations is

\(\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}\)

Quantity A:

\(\displaystyle C(10,5)=\frac{10!}{(10-5)!5!}=252\)

Quantity B:

\(\displaystyle C(20,2)=\frac{20!}{(20-2)!2!}=190\)

Quantity A is greater.

Since in this problem the order of selection does not matter, we're dealing with combinations.

With \(\displaystyle k\) selections made from \(\displaystyle n!\) potential options, the total number of possible combinations is

\(\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}\)

In terms of finding the maximum number of combinations, the value of \(\displaystyle k\) should be 

\(\displaystyle k_{max}=\frac{n_{even}}{2};k_{max}=\frac{n_{odd}\pm 1}{2}\)

Since there are twelve options, a selection of six scoops will give the maximum number of combinations.

\(\displaystyle C(12,6)=\frac{12!}{(12-6)!6!}=924\)

\(\displaystyle C(12,7)=792\)

\(\displaystyle C(12,8)=495\)

\(\displaystyle C(12,9)=220\)

\(\displaystyle C(12,10)=66\)

Step 1: We need to read the question carefully. Order does not matter here.

Step 2: Order does not matter, so we need to use Combination.

Step 3: The combination formula is \(\displaystyle nCr=\frac{n!}{r!(n-r)!}\).

Step 4: We need to find the value of \(\displaystyle n\) and \(\displaystyle r\).

The value of \(\displaystyle n\) is how many players the coach can choose from, so \(\displaystyle n=11\). 
The value of \(\displaystyle r\) is how many players that the coach can choose at one time, so \(\displaystyle r=5\).

Step 5: Plug in the values of n and r into the equation in step 2:

\(\displaystyle nCr=\frac{11!}{5!(11-5)!}\)

Step 6. Simplify the equation in step 5. The "!" means that I multiply that number by every other number below until 1.

\(\displaystyle nCr=\frac{11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{5\cdot 4\cdot 3\cdot 2\cdot 1\cdot 6 \cdot5\cdot 4\cdot 3\cdot 2\cdot 1}\)

Step 7: Cross out any terms that are on both the top and the bottom. We see \(\displaystyle 1-6\) is on top and bottom.

\(\displaystyle nCr=\frac{11\cdot 10\cdot 9\cdot 8\cdot 7}{5\cdot4\cdot3\cdot2\cdot1}\)

Step 8: Cross out \(\displaystyle 5\cdot2\) in the denominator with \(\displaystyle 10\) in the numerator. Rewrite.

\(\displaystyle nCr=\frac{11\cdot 9\cdot 8\cdot 7}{4\cdot 3\cdot 1}\)

Step 9: Divide \(\displaystyle 8\) in the numerator by \(\displaystyle 4\) in the denominator.

\(\displaystyle nCr=\frac{11\cdot 9\cdot 2\cdot 7}{3\cdot 1}\)

Step 10: Divide \(\displaystyle 9\) in the numerator by \(\displaystyle 3\) in the denominator.

\(\displaystyle nCr=11\cdot 3\cdot2\cdot7\)

Step 11: Multiply the right side

\(\displaystyle 11C5=462\)

There are 462 ways that the coach can choose 5 players out of 11 players on the bench.

Step 1: Read the question carefully. Look for hints of restrictions..

There are no order in which players can be chosen, which goes against the definition of Permutation. Permutation is the arrangement of objects by way of order.. If it's not permutation, it's Combination.

Step 2: Write what we know down..

Total Players \(\displaystyle (n)\)=\(\displaystyle 13\)
Choosing # of players \(\displaystyle (r)\)=\(\displaystyle 6\)..

Step 3: Plug in the numbers to the formula: \(\displaystyle nCr\)..

We ger 13C6. 

There is no need to evaluate this expression...

There are two types of statistical calculations that are used when dealing with ordering a number of objects. When the order does not matter it is known as a combination and denoted by a C.

\(\displaystyle C(n,k)=\frac{n!}{k!(n-k)!}\)

Thus the formula for this particular combination is,

\(\displaystyle C (8,5) = P \frac{(8,5)}{5!}\)

\(\displaystyle \frac{8\times 7\times 6\times 5\times 4}{5\times 4\times 3\times 2\times 1} =\)

The \(\displaystyle 5 \times 4\) will cancel out because it is in the numerator and denominator,

\(\displaystyle \frac{8\times 7\times 6}{3\times 2\times 1} = \frac{336}{6} = 56\).