Introduction Of Combinations - Practice Questions & MCQ

So far our task was always to “arrange” objects i.e. to place them in a specific order among themselves.

Sometimes we would be interested in only “selecting” a few objects out of the given objects. In this case, we just need to “select” and we do not need to “arrange” them in an order. 

E.g., we need to select 4 students out of the 15 students who will represent the college at a quiz or we need to form an academic committee of 3 professors from 10 professors. In this case, who is selected “first”, who is selected “second” and so on does not matter. The words “first” and “second” implicitly imply an “ordering”. What matters in the case of selection is only the composition of the final “group”. 

The notation of selecting r objects from n given object is ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}$. Let’s derive the value of ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}$., and its relation with permutation notation.

Let's say we want to arrange 2 objects out of 5 objects: A, B, C, D, and E then using the concept of permutation we can do this in  ways.

We can calculate the same thing by another method: by selecting 2 things out of 5, which can be done as  , and then arranging the 2 selected things which can be done in 2! ways. So we have 

$$
\begin{aligned}
& { }^5 \mathrm{C}_2 \times 2!={ }^5 \mathrm{P}_2 \\
& { }^5 \mathrm{C}_2=\frac{{ }^5 \mathrm{P}_2}{2!} \\
& { }^5 \mathrm{C}_2=\frac{5!}{(5-2)!2!}=\frac{5!}{3!2!}
\end{aligned}
$$

We can generalize this concept for $r$ object to be selected from given $n$ objects as
$$
\begin{aligned}
& { }^n C_r \times r!={ }^n P_r \\
& { }^n C_r=\frac{{ }^n P_r}{r!} \\
& { }^n C_r=\frac{n!}{(n-r)!r!}
\end{aligned}
$$

Where 0 ≤ r ≤ n, and r is a whole number.

Now we have the value of 

Example: In ICC World Cup 2019 total of 10 teams participated and each team had to play one game in the league stage with all other teams before qualifying for the semifinals, so how many total games will be played in the league stage?

Solution: For playing a game we need to select two teams. So this is a simple problem of selecting two teams, so this can be done in 

$\begin{aligned} & { }^{10} \mathrm{C}_2=\frac{10!}{(10-2)!2!} \\ & \frac{10 \times 9 \times 8!}{8!\times 2!}=\frac{10 \times 9}{2}=45\end{aligned}$

Hence in total 45 games will be played in the league stage.