# Examples - Combinations as Selections

Examples - Combinations as Selections
Go back to  'Permutations and Combinations'

Example 1. A cricket team has a squad of 15 members.

(a) In how many ways can a playing team of 11 members be selected?

(b) In how many ways can a batting order consisting of 11 players be made?

(c) In how many ways can a playing team be selected if one particular player must be chosen?

(d) In how many ways can a batting order consisting of 11 players be made, if one particular player must not be chosen?

Solution:

(a) In this part, you only have to select 11 players out of 15. This is therefore a combination problem. The answer is $$^{15}{C_{11}}$$.

(b) Now, not only do you have to select a group of 11 players, but you also have to decide their batting order, or arrangement. This is a permutation problem, and the answer will be   $$^{15}{P_{11}}$$.

(c) If we must have one particular player in our team, then we need to choose 10 players out of the remaining 14. This can be done in $$^{14}{C_{10}}$$ ways.

(d) One particular player must not be chosen. From the remaining 14, we need to determine the number of all permutations or arrangements of 11 players. The answer is $$^{14}{P_{11}}$$.

Example 2: In a party of 10 people, each person shakes hands with every other person. How many handshakes are there?

Solution: Each unique handshake corresponds to a unique pair of persons. Also, note that the order of the two people in the pair does not matter. For example, if X and Y are two people, then XY and YX won’t be counted separately; only the pair {X,Y} will be counted.

Thus, we need to find the number of ways in which 2 people can be selected out of 10. Clearly, the answer is

$^{10}{C_2} = \frac{{10!}}{{2!\left( {10 - 2} \right)!}} = \frac{{10!}}{{2!8!}} = 45$

Make sure you understand this calculation.

Example 3: A class consists of 10 boys and 10 girls. A group of 5 students needs to be selected from this class. In how many ways can this be done if

(a) there is no constraint on the composition of the group?

(b) there is at least one boy in the group?

(c) there are 2 boys and 3 girls in the group?

(d) there are at least 2 girls and 2 boys in the group?

Solution:

(a) In this case, we simply have to select any 5 students of 20, which can be done in $$^{20}{C_5}$$ ways.

(b) First, we count all the groups in which there are no boys – all the members of the groups are girls. The total number of girls is 10. Selecting 5 out of 10 girls can be done in $$^{10}{C_5}$$ ways. Now think about this carefully: the number of all possible groups of size 5 is $$^{20}{C_5}$$, and out of these, $$^{10}{C_5}$$ groups consist of only girls. Thus, the number of groups which consist of at least one boy is $$^{20}{C_5}{ - ^{10}}{C_5}$$.

(c) We first select 2 boys out of 10. This can be done in $$^{10}{C_2}$$ ways. Next, we select 3 girls out of 10 – this can be done in $$^{10}{C_3}$$ ways. For each of the $$^{10}{C_2}$$ selections of boys, there are $$^{10}{C_3}$$ selections of girls. Thus, using the FPC, the total number of ways of forming a group consisting of 2 boys + 3 girls is $$^{10}{C_2}{ \times ^{10}}{C_3}$$.

(d) The constraint “at least 2 girls and 2 boys” can be divided into two separate cases: (i) 2 boys + 3 girls (ii) 3 boys + 2 girls. We have already calculated the number of groups consisting of 2 boys + 3 girls: $$^{10}{C_2}{ \times ^{10}}{C_3}$$. Similarly, the number of groups consisting of 3 boys + 2 girls will be $$^{10}{C_3}{ \times ^{10}}{C_2}$$ (the same as the first case). Thus, the number of groups consisting of at least 2 girls and 2 boys is

$\left( {^{10}{C_2}{ \times ^{10}}{C_3}} \right) + \left( {^{10}{C_3}{ \times ^{10}}{C_2}} \right) = 2{ \times ^{10}}{C_2}{ \times ^{10}}{C_3}$

Example 4:From a group of 6 boys and 4 girls, a selection of 5 children needs to be made. In how many ways can this be done if the selection has

(a) at least 2 girls?

(b) at the most 2 girls?

Solution:  (a) We divide the situation of “at least 2 girls” into the following cases, and count the number of groups corresponding to each case:

 Case Num. of girls Num. of boys Num. of possible selections 1 2 3 $^4{C_2}{ \times ^6}{C_3} = 120$ 2 3 2 $^4{C_3}{ \times ^6}{C_2} = 60$ 3 4 1 $^4{C_4}{ \times ^6}{C_1} = 6$

Thus, the total number of selections consisting of at least 2 girls is 120 + 60 + 6 = 186.

(b) This part is left to you as an exercise. The answer will be 186 again.

Example 5: A class has 25 students. For a school event, 10 students need to be chosen from this class. 4 of the students of the class decide that either four all of them will participate in the event, or none of them will participate. In how many ways can the selection of 10 students be done?

Solution:  Given the specified constraint, we divide the set of all possible selections of 10 students into two groups:

1. The selections which include the 4 students; we already have 4 students – we need to select 6 more students out of remaining 21 students. This can be done in $$^{21}{C_6}$$ ways. Thus, the number of possible selections which include the 4 students is $$^{21}{C_6}$$.
2. The selections which do not include the 4 students; now we need to select 10 students out of the remaining 21. This can be done in $$^{21}{C_{10}}$$ ways. Thus, the number of possible selections which do not include the 4 students is $$^{21}{C_{10}}$$.

The total number of selections possible under the specified constraint is $$^{21}{C_6}{ + ^{21}}{C_{10}}$$.

Example 6: How many diagonals are there in a polygon with n sides?

Solution: If we select any two vertices of the polygon and join them, we will get either a diagonal or a side of the polygon. The number of ways of selecting two vertices out of n is $$^n{C_2} = \frac{{n\left( {n - 1} \right)}}{2}$$. Out of these selections, n correspond to the sides of the polygon. Thus, the number of diagonals is:

$\frac{{n\left( {n - 1} \right)}}{2} - n = \frac{{n\left( {n - 3} \right)}}{2}$

Combinatorics
Combinatorics
grade 9 | Questions Set 2
Combinatorics