Hi. This lesson will cover a few examples relating to combinations.

**Example 1 **A question paper consists of 10 questions of which a student needs to answer any 7. In how many ways can the student make his selection?

**Solution **This is a simple case of selection of 7 objects (questions) out of 10 distinct objects. The number of ways will be ^{10}C_{7} = 120

**
Example 2 **Find the number of ways in which a team of 4 can be selected from a group of 10 people A

_{1}, A

_{2}, A

_{3}, …, A

_{10}such that

(i) there are no restrictions

(ii) A_{1} and A_{3} must be selected

(iii) A_{2} must not be selected

(iv) if A_{6} is selected then A_{4} must also be selected

(v) if A_{5} is selected then A_{7} must not be selected

(vi) A_{9} is selected if and only if a A_{10 }is selected

**Solution **Let’s take ‘em one by one!

(i) This one’s plain: ^{10}C_{4}

(ii) In this case, we must select A_{1} and A_{3}. Done. Now, out of the remaining 8, we’ve to select 2 more. The number of selections will be ^{8}C_{2}

(iii) In this case, we’ll remove A_{2} from the group, and select any 4 from the remaining group of 9 people – in ** ^{9}C_{4}** ways.

(iv) This one is a little complicated. We’ll have to divide our counting into 2 cases. First in which A_{6} is selected, and the second in which A_{6} is not selected.

**Case 1:** According to the restriction, if A_{6} is selected then A_{4} also must be selected. So we’ll select these 2, plus 2 more from the remaining 8, in ^{8}C_{2} ways.

**Case 2:** If A_{6} is not selected, then we can select any 4 out of the remaining 9, in ^{9}C_{4} ways.

Since it will be **either** Case 1 **or** Case 2, the number of ways in both these cases will be added (the addition principle). The total possible selections will be ** ^{8}C_{2} + ^{9}C_{4}**.

(v) Again, we’ll divide our counting into 2 cases. First in which A_{5} is selected, and the second in which A_{5} is not selected.

**Case 1:** According to the restriction, if A_{5} is selected then A_{7} must not be selected. So we’ll remove A_{7}, and select 3 more from the remaining 8: ^{8}C_{3} ways.

**Case 2:** If A_{5} is not selected, then we can select any 4 out of the remaining 9, in ^{9}C_{4} ways.

The total possible selections will be ** ^{8}C_{3} + ^{9}C_{4}**.

(vi) In this case, either both A_{9} and A_{10} will be selected, or both will not be selected.

Case 1: Select A_{9} and A_{10}, and select 2 more from the remaining 8: ^{8}C_{2} ways

Case 2: Remove A_{9} and A_{10} from the group, and select 4 from the remaining 8: ^{8}C_{4}ways

Total number of ways: ^{8}C_{2} + ^{8}C_{4}

Let’s now complicate things a bit more..

**Example 3 **In how many ways can a committee of 8 be selected from a group of 10 men and 12 women, such that, in the committee,

(i) there are 3 men and 5 women;

(ii) there are atleast 6 women;

(iii) there is atleast one man;

(iv) there are atmost 2 women;

(v) there are more men than women ?

**Solution **In this problem, we’ll divide our task of selection into 2 subtasks – (a) select the men and (b) select the women. Since both of these need to be completed to complete the selection, we’ll multiply the number of ways obtained in both these cases (the multiplication principle)

(i) The men can be selected in ^{10}C_{3} ways, and the women can be selected in ^{12}C_{5} ways. Therefore the total number of ways will be ** ^{10}C_{3} x ^{12}C_{5}**.

(ii) Now in this case, there can either be 6 women (and 2 men), or 7 women or all 8 women.

We’ll therefore count the three cases separately and add them together. Similar to (i) above, when there are 6 women (and 2 men), the number of cases will be ^{10}C_{2} X ^{12}C_{6}. And when there are 7 women (and 1 man) the number of cases will be ^{10}C_{1} x ^{12}C_{7}. Finally in the case of 8 women and 0 men, the number of selections will be ^{10}C_{0} x ^{12}C_{8}, or simply ^{12}C_{8}. The final answer becomes ^{10}C_{2} x ^{12}C_{6} + ^{10}C_{1} x ^{12}C_{7} + ^{12}C_{8}

(iii) This could be counted in a similar manner as (ii) above. But there is a better method. If from all the possible selections, we remove those in which no man is selected (or all-woman committees), we’ll be left with those cases in which there is atleast one man. Sounds right?

The total number of 8-member committees will be ^{22}C_{8}. The total number of all-women committees will be ^{12}C_{8}. Therefore our required answer will be ** ^{22}C_{8} – ^{12}C_{8}**. Pretty neat!

(iv) Either all men, or 7 men and 1 woman, or 6 men and 2 women: ^{10}C_{8} + ^{10}C_{7} x ^{12}C_{1} + ^{10}C_{6} x ^{12}C_{2}

(v) I’ll leave this one for you as an exercise.

More examples to follow in the next lesson. See you there!

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