4.6 Counting rules - university of calgary - base content - v2015  (Page 3/6)

When only selecting two pictures to hang, the outcomes { ABCD, ABDC } are considered the same because picture A is in the first position and picture B is in the second position for both outcomes. We don’t care about the arrangement of the last two pictures C and D because they were not chosen to be hung. The outcome { BACD, BADC } are considered the same because picture B is in the first position and picture A is in the second position. Since the order matters, having picture A in the first position and B in the second is a different arrangement from having B in the first position and A in the second position. This is why in the formula for permutation we are dividing the sample space for 4! by two.

Earlier in the text we discussed random samples and the use of a random generating table to select a random sample. How many random samples are possible? We can use Combinatorial mathematics to determine the number of possible samples. Suppose we have our four pictures and we will randomly choose two to give to a friend. In this example, the order of the two pictures that we select and the order of the two we have not selected does not matter.

{ ABCD, ABDC, BACD, BADC } are considered the same because picture A and B are in the first two positions (pictures you will give your friend) and C and D are the pictures you will keep (pictures in third and fourth position).

We take the original sample for 4! = 24, divide it by two (possible arrangements of the two pictures selected) and divide this by two (possible arrangements of the two pictures not selected). We are left with six possible random samples. The possible random samples of two pictures you can give your friend are AB, AC, AD, BC, BD, and CD .

$\frac{4!}{2!2!}=$ $\frac{4\times 3\times 2\times 1}{(2\times 1)(2\times 1)}=6$

Combination

Combination is selecting a sample consisting of r elements from n distinct available items. Below are different ways that a combination can be represented.Insert paragraph text here.

For the picture example, there are 4 available items ( n = 4) and we are going to select two ( r = 2). We are not concerned about the order of the pictures selected nor the order of the pictures not selected.

$\left(\begin{array}{c}4\\ 2\end{array}\right)=$ C_2^4= ${}_4{C}_2=$ $\frac{4!}{2!(4-2)!}=$ $\frac{4!}{2!2!}=$ $\frac{4\times 3\times 2\times 1}{(2\times 1)(2\times 1)}=6$

Formula review

Permutation : P_r^n={}_n{P}_r=\frac{n!}{(n-r)!}=n(n-1)n(n-2)\mathrm{...}(n-r+1)

Combination : $\left(\begin{array}{c}n\\ r\end{array}\right)=C_r^n={}_n{C}_r=\frac{n!}{r!(n-r)!}$