Combinations: Examples


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 10C7 = 120


Example 2 
Find the number of ways in which a team of 4 can be selected from a group of 10 people A1, A2, A3, …, A10 such that

(i) there are no restrictions
(ii) A1 and A3 must be selected
(iii) A2 must not be selected
(iv) if A6 is selected then A4 must also be selected
(v) if A5 is selected then A7 must not be selected
(vi) A9 is selected if and only if a A10 ­is selected

Solution Let’s take ‘em one by one!

(i) This one’s plain: 10C4

(ii) In this case, we must select A1 and A3. Done. Now, out of the remaining 8, we’ve to select 2 more. The number of selections will be 8C2

permutations combinations examples

(iii) In this case, we’ll remove A2 from the group, and select any 4 from the remaining group of 9 people – in 9C4 ways.

permutations combinations examples

(iv) This one is a little complicated. We’ll have to divide our counting into 2 cases. First in which A6 is selected, and the second in which A6 is not selected.

Case 1: According to the restriction, if A6 is selected then A4 also must be selected. So we’ll select these 2, plus 2 more from the remaining 8, in 8C2 ways.

Case 2: If A6 is not selected, then we can select any 4 out of the remaining 9, in 9C4 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 8C2 + 9C4.

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(v) Again, we’ll divide our counting into 2 cases. First in which A5 is selected, and the second in which A5 is not selected.

Case 1: According to the restriction, if A5 is selected then A7 must not be selected. So we’ll remove A7, and select 3 more from the remaining 8: 8C3 ways.

Case 2: If A5 is not selected, then we can select any 4 out of the remaining 9, in 9C4 ways.

The total possible selections will be 8C3 + 9C4.

permutations combinations examples

(vi) In this case, either both A9 and A10 will be selected, or both will not be selected.

Case 1: Select A9 and A10, and select 2 more from the remaining 8: 8C2 ways

Case 2: Remove A9 and A10 from the group, and select 4 from the remaining 8: 8C4ways

Total number of ways: 8C2 + 8C4

permutations combinations examples

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 10C3 ways, and the women can be selected in 12C5 ways. Therefore the total number of ways will be 10C3 x 12C5.

(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 10C2 X 12C6. And when there are 7 women (and 1 man) the number of cases will be 10C1 x 12C7. Finally in the case of 8 women and 0 men, the number of selections will be 10C0 x 12C8, or simply 12C8. The final answer becomes 10C2 x 12C6 + 10C1 x 12C7 + 12C8

(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 22C8. The total number of all-women committees will be 12C8. Therefore our required answer will be 22C812C8. Pretty neat!

(iv) Either all men, or 7 men and 1 woman, or 6 men and 2 women: 10C8 + 10C7 x 12C1 + 10C6 x 12C2

(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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