0.10 Sets and counting  (Page 8/9)

We have the following four combinations.

ABC BCD CDA BDA

Since every combination has three letters, there are $3!$ permutations for every combination. We list them below.

ABC BCD CDA BDA

ACB BDC CAD BAD

BAC CDB DAC DAB

BCA CBD DCA DBA

CAB DCB ACD ADB

CBA DBC ADC ABD

The number of permutations are $3!$ times the number of combinations. That is

or $\mathrm{4C3}=\frac{\mathrm{4P3}}{3!}$

In general, $\text{nCr}=\frac{n\text{Pr}}{r!}$

Since $n\text{Pr}=\frac{n!}{\left(n-r\right)!}$

We have, $\text{nCr}=\frac{n!}{\left(n-r\right)!r!}$

We use the above formula.

Since the order is not important, it is a combination problem, and the answer is

$\mathrm{7C5}=\text{21}$ .

Since the line that goes from point $A$ to point $B$ is same as the one that goes from $B$ to $A$ , this is a combination problem.

It is a combination of 6 objects taken 2 at a time. Therefore, the answer is

Note that between any two people there is only one hand shake. Therefore, we have

Let us suppose the taxi driver drives from the point $A$ , the lower left hand corner, to the point $B$ , the upper right hand corner as shown in the figure below.

To reach his destination, he has to travel ten blocks; five horizontal, and five vertical. So if out of the ten blocks he chooses any five horizontal, the other five will have to be the vertical blocks, and vice versa. Therefore, all he has to do is to choose 5 out of ten.

The answer is 10C5, or 252.

Alternately, the problem can be solved by permutations with similar elements.

The taxi driver's route consists of five horizontal and five vertical blocks. If we call a horizontal block $H$ , and a vertical block a $V$ , then one possible route may be as follows.

Clearly there are $\frac{\text{10}!}{5!5!}=\text{252}\text{ permutations}$ .

Further note that by definition $\text{10}\mathrm{C5}=\frac{\text{10}!}{5!5!}$ .