Combinations Examples - Concepts & Word Problems

# Combinations Examples

In real life, on many occasions, we need to deal with a number of items at the same time.  And we need to decide how may among them to be considered for a given purpose. For example, we have to select $11$ players for a football team. If there are only $11$ persons then there is no difficulty, all the $11$ persons are considered. But, say, if we have $15$ qualified persons for the game, then there are a number of ways by which the team of $11$ players can be formed. A selector should work out the different combinations to study the pros and cons of each combination and take a prudent decision. Hence the concept of combination is very important and it is widely applied in certain probability problems.

## Combination Concepts

The technique of finding the number of combinations from a given set is based on 'counting principle' which is just a simple logic. Let us elaborate with a simple case. Consider three items $A, B$ and $C$. Out of these $3$ words how many pairs can be selected? With our common sense we can say, if the first letter is assumed as $A$, there are two possibilities for the second letter. It could be either $B$ or $C$ and hence the possible pairs are $AB$ and $AC$. Extending the same logic, we can form $BC, BA$ and $CA$ and $CB$ by first keeping the letter $B$ and letter $C$ respectively. Thus, there is a total of $6$ arrangements. Note that we are not using the word combinations. Let us see why. We have considered $AB$ and $BA$ are two different arrangements because of the order of the letters though the same letters are used. But as a combination, both the arrangements are same and same is the case with the pairs $AC, CA$ and  $BC, CB$. Hence we end up with the total number of combinations as $3$. However, it may be noted that in certain cases the order of the items do matter, for example a password $AB$ will not work if entered as $BA$. In such cases the arrangements are called as permutations. In the same example we can say that the total number of permutations is $6$.

When the number of items is large it may be cumbersome to find the required number of combinations by using the counting principle. Using the same concept, mathematicians have derived a formula which is easier to find the number of combinations in such cases. If there are '$n$' items, the number of combinations taking '$r$' items at a time is denoted as $^nC_r$ and the formula is,

$^nC_r$ = $\frac{n!}{(n -r)!r!}$

## Examples

Example 1:

From a total number of $30$ members a committee of $5$ persons to be formed. In how many ways the committee can be formed?

Solution:

A committee just requires the required number of people. It does not matter who is the picked up first and who is picked up last. Hence the order in selecting the people is not important and hence it is a question of combination.

In this case, it is the number of combination of $30$ persons choosing $5$ persons at a time and hence it is,

$^{30}C_5$ = $\frac{30!}{(25)!5!}$ = $142506$
Example 2:

A football team of $11$ players is to be selected from $15$ eligible players, which includes the captain and the only available goal keeper. In how many ways the team can be selected?

Solution:

This is also a combination problem but attached some conditions attached. In any combination of the teams, the captain and the goal keeper must be included. In other words, the selection now becomes as the combination of remaining $13$ players with only $9$ players to be considered now. Therefore, the possible number of ways the team can be finalized is,

$^{13}C_9$ = $\frac{13!}{(4)!9!}$ = $715$
Example 3:

A board of $5$ directors with $3$ men and $2$ women is to be formed from the council of $8$ men and $5$ women. In how many ways the board can be formed?

Solution:

This problem is not only a conditional combination problem but on a closer study it reveals that it is sum of two different combinations. The first is choosing $3$ men out of $8$ men and choosing $2$ women out of $5$ women.

The number of combinations in the first case is $^8C_3$ and in the second case is $^5C_2$. But it may be realized for each of the $^8C_3$ combinations there are $^5C_2$ combinations.

Therefore, effectively the number of all possible combinations is,

$^8C_3\ \times\ ^5C_2$ = $56 \times 10$ = $560$.

Thus the desired form of board can be formed in $560$ ways.
Example 4:

From a standard pack of cards $4$ cards are drawn. What is the probability that the set contains at least two red cards?

Solution:

In a standard pack the total number of cards is $52, 26$ each in red and black colors. The condition 'at least two red cards' means that the set of $4$ cards could be with $2$ red and $2$ black, $3$ red and $1$ black or all $4$ red. So, the final number of combinations is sum of all these combinations. And in each case it is a product of combinations of red and the combinations of black cards. Therefore, the final number of combinations is,

$^{26}C_2\ \times\ ^{26}C_2\ +\ ^{26}C_3\ \times\ ^{26}C_1\ +\ ^{26}C_4\ \times\ ^{26}C_0$ = $325 \times 325 + 2600 \times 26 + 14950 \times 1$ = $188175$

Now the number of all possible combinations of drawing $4$ cards from $52$ cards (sample space) is,

$^{52}C_4$  = $270725$

Therefore, the required probability is $(\frac{188175}{270725})$ $\approx\ 0.695$
Example 5:

In the following grid, in how many different ways you can reach from $A$ to $B$?

Solution:

The condition is that you can only go along the side of any square and you can move only towards up or towards right.

This might look very hard and brain teasing. But the problem can easily be solved by using principle of combinations. The displacement of point $B$ is $7$ units horizontal and $6$ units vertical from point $A$. Thus, in any path you need to cover $13$ units. Therefore, if you find the number of paths to cover the horizontal displacement, the vertical paths are fixed by themselves. Alternately, if you find the number of paths to cover the vertical displacement, the horizontal paths are fixed by themselves. Thus, the required answer is either $^{13}C_7$ or $^{13}C_6$ (both are same! and equal to $1716$).

So, there are $1716$ ways by which you can reach point $B$ from point $A$.