Applications Of Selections - Practice Questions & MCQ

Let us take an example of selecting things from two or more different groups:

Out of 5 men and 6 women in how many ways can a committee of 5 members be selected such that at least 2 members are women? 

Solution:

Following cases are possible for at least 2 women,

2 women +3 men $={ }^6 \mathrm{C}_2 \times{ }^5 \mathrm{C}_3$
3 women +2 men $={ }^6 \mathrm{C}_3 \times{ }^5 \mathrm{C}_2$
4 women +1 men $={ }^6 \mathrm{C}_4 \times{ }^5 \mathrm{C}_1$
5 women $={ }^6 \mathrm{C}_5$
So, the total number of ways $={ }^6 \mathrm{C}_2 \times{ }^5 \mathrm{C}_3+{ }^6 \mathrm{C}_3 \times{ }^5 \mathrm{C}_2+{ }^6 \mathrm{C}_4 \times{ }^5 \mathrm{C}_1+{ }^6 \mathrm{C}_5=431$

Restricted Combination

The number of selection of r objects from n different objects:

1. When k particular things are always included  = ${ }^{n-k} C_{r-k}$

This can be comprehended as taking out those k things that have to be included which can be done in 1 way and then finding the ways in which r-k objects can be selected from remaining n - k things, and putting those k things (which are already taken out) in r-k selected objects.

2. k particular things are never included  = ${ }^{n-k} C_r$

This can be comprehended as taking out k things that are not to be selected which can be done in 1 way and then finding the ways of selecting r things from n-k things.

3. The number of ways selecting r things out of n different things such that p particular objects are always included and q particular objects are always excluded = ${ }^{n-p-q} C_{r-p}$

This can be comprehended as taking out the q objects that should not be selected and putting it out and then taking out p objects that have to be selected and then finding ways of selecting r-p objects out of n- p-q objects and putting back p objects in r-p selected objects.

Example: In how many ways a cricket team can be selected out of 16 players such that 5 certain players must be included in the team?

Solution: Since 5 certain players have to be included so be need to select 11-5 = 6 players from 16 - 5 = 11 players.

      So we can select the team in ${ }^{16-5} \mathrm{C}_{11-5}={ }^{11} \mathrm{C}_6=\frac{11!}{5!6!}$

If there are n points in the plane  and out of which no three are collinear then,

1. Total No. of lines that can be formed using these n points = ⁿC₂

2. Total No. of triangles that can be formed using these n points = ⁿC₃

3. Total no. of Diagonals that can be formed in n sided polygon = ⁿC₂ - n

If there are n points in the plane and out of which m points are collinear, then,

1. The total No. of different lines that can be formed by joining these $n$ points is ${ }^n C_2-{ }^m C_2+1$
2. Total No. of different triangles that can be formed by joining these n points is ${ }^{\mathrm{n}} \mathrm{C}_3-{ }^{\mathrm{m}} \mathrm{C}_3$
3. The total No. of different quadrilaterals formed by joining these $n$ points is
$$
{ }^{\mathrm{n}} \mathrm{C}_4-\left({ }^{\mathrm{m}} \mathrm{C}_3 \cdot{ }^{\mathrm{n}-\mathrm{m}} \mathrm{C}_1+{ }^{\mathrm{m}} \mathrm{C}_4\right)
$$

Number of Parallelograms

If m parallel lines in a plane are intersected by the family of other n parallel lines, then the total number of parallelograms formed is 

$$
{ }^{\mathrm{m}} \mathrm{C}_2 \cdot{ }^{\mathrm{n}} \mathrm{C}_2=\frac{\mathrm{mn}(\mathrm{~m}-1)(\mathrm{n}-1)}{4}
$$

Number of rectangles and squares
$$
\sum_i^n r^3
$$
1. Number of rectangles of any size in a square of size n x n is $r=1$ and number of squares of any size is $r=1$.
2. In a rectangle of size $\mathrm{n} \times \mathrm{p}(\mathrm{n}<\mathrm{p})$ number of rectangles of any size is $\frac{n p}{4}(n+1)(p+1)$.

To determine the number of ways to reach in the shortest way from point A to B.

When considering the possible paths or shortest path one can observe that the total number of steps in the forward direction is 6-R(Right) and in the upward direction is 4-U(Upward)

Now, If we arrange these 6 Rs and 4 Us in any way, it comes out to be a shortest path.

Or one can say that first find all the possible steps and arrange them to get the total number of possible ways.

Using "u" and "r" we can write out a path:

r r r r r r u u u u

r r r u u u u r r r

and others......

Hence, the total number of ways is  or,