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Fri Sep 18, 2020

In this post, we will find out about logical reasoning concepts on determination and gathering arrangement that is habitually asked in CAT exam. This point for the most part manages the choice of a group of state 'r' individuals from 'n' (n>r) accessible for choice or it very well may be the choice of panel of certain number of individuals. Certain number of imperatives drives this determination. So as to comprehend these compels and the understood subtleties identified with them, let us start the conversation with an example.

Question: Among five students of group I – A, B, C, D, E and six students of group II – U, V, W, X, Y, Z, a team of five students is selected such that it consists exactly three students from group II. It is also known that:

C and V cannot be selected together.

If B is selected, neither U nor V can be selected.

Among A,D,E and Y exactly two persons are to be selected.

If E is in the team, at most one among U and W can be in the team.

If A is selected, X has to be selected.

Z will be in the team if and only if C is selected.

Find the total number of such teams possible.

**Approach to solve:**

Let us take each condition one by one and try to get the obvious and hidden details about each information available.

1. The team consists of exactly three students from group II:

We know that the number of students in the team is five. If exactly three students from group II is in the team then the rest two students must come from group I. This will always be the distribution in the team – two from group I and three from group II.

2. C and V cannot be selected together:

I.-If C is in the team, V cannot be in the team.

II.-If V is in the team, C cannot be in the team.

III.-It does not mean that any one between C and V need to be selected. Both of them could simultaneously be out of the team

3. If B is selected, neither U nor V can be selected:

I.-If B is in the team, U or V is not in the team.

II.-If U or V or both are in the team, then B is not in the team.

III.-All 3 can simultaneously be out of the team.

4. Among A, D, E and Y, exactly two persons are to be selected.

I.-We have to select any two among A, D, E and Y, it is mandatory.

II.-It can be – (A,D), (A,E), (A,Y), (D,E), (D,Y), (E,Y)

5. If E is in the team, at most one among U and W can be in the team.

I.-E and U can be in the team together.

II.-E and W can be in the team together.

III.-E can be in the team without U or W in the team.

IV.-E, U and W cannot be in the team together.

V.-U and W can be in the team without E being in the team.

VI.-All three can simultaneously be out of the team.

6. If A is selected, X has to be selected.

I.- If A is in the team, X has to be in the team.

II.- If X is in the team, A may or may not be in the team.

III.- If A is not in the team, X may or may not be in the team.

IV.- If X is not in the team, A will not be in the team.

7. Z will be in the team if and only if C is selected

I.- This means if Z is in the team C is in the team

II.- If C is in the team, Z is also in the team.

III.- Both will simultaneously be in the team or out of the team.

‘If and Only If’ is the most important sentence out of as it implies that either select both or none, thus making them a pair.

Checking all the possibilities let us try to make all the possible teams.

**Where to start:**

There can be a number of ways to start the puzzle. Three of the easiest approaches are:

1. Select two members from group I.

Group I has 5 members and we need to select 2 members from group I so total possible cases can be 5C2 = 10. There is a possibility that not all of the 10 cases will result in a team and we have to solve individually for all 10 cases. For instance, we select B and C as two members from group I but from condition 3 we know that at least one among A, D or E (all three belongs to group 1) would be in the team thus making this case useless.

2. Select three members from group II

Selecting three from a group of 6 is 6C3 = 20. Solving 20 cases would be even more time consuming.

3. Select two from condition 3.

We know that among A, D, E and Y, exactly two will be in the team. Number of possible cases are six as already discussed before. Solving 6 cases would definitely be less time consuming than other approaches. In addition, since we already abiding to one of the conditions so now we need to check for only five remaining conditions.

Let us start making teams by selecting two members as per condition 3.

**Teams: **

1.- A and D are selected in the team (from condition 3), which means that the two members of group I are already selected and we need to select 3 members from group II. From condition 5, we can say that X will be in the team (A is selected then X is also selected). Y and Z cannot be selected (Y cannot be selected because among A, D, E and Y only 2 can be in the team and we already selected A and D; Z cannot be selected as C is not in the team (condition 6)). We are left with U, V and W. We need to select two among U, V and W so we can have three cases.

2.- Next we select A and E from A. D. E and Y. A and E belong to group I and so we need three more from group II. X will be one of them (condition 5). Now, since E is in the team either one of U or W can be in the team (condition 4). Y and Z cannot be in the team because of the reason explained in previous point. Therefore, we have two options. U and V or W and V.

3.- Now, we select A and Y. A belongs to group I and Y belongs to group II. We need one more member from group I and two more from group II. From group I, we can select B or C (D or E cannot be selected because of condition 3).

With B, we cannot have either U or V in the team (condition 2) and since A is in the team, X has to be in the team. Without C, Z cannot be in the team. The only option left is W.

With C, Z will also be in the team (condition 6). Thus, all 5 are selected and no more case possible.

4.- Next we select D and E. Both are from group I, leaving space for 3 group II members. Condition 4 allows only one among U or W because E is in the team. Again, Y and Z cannot be in the team (explained earlier). Thus, the only option left is with V and X.

Note that X can be in the team even if A is not in the team.

.

5. Next, we select D and Y. D is from group I and Y from group II. One more from group I can be B or C. Similar to the third case when we selected A and Y but here we are not restricted to the selection of X alone as there is no A. X will be in some of the teams but not in all teams. With C in the team, Z will also be in the team but V cannot be in the team (condition 1). With B in the team, neither U nor V can be in the team so we should select X and W.

6. Last case is with E and Y selection. With E in the team, either U or W can be in the team. From group I we can select either B or C. Rest of the case is similar to the last case.

Did we miss any case? No. How can we be so sure? Because we discussed all the 6 cases coming out of condition number 3. Any team would include exactly two among A, D, E, Y and we made a case for all possible combinations. This is a much better practice, as it would avoid any confusion related to number of cases possible. We should always check subcases for further combinations with each member one by one. Following a flow diagram surely helps in determining the combinations with given restrictions. When there is no such condition which gives us any concrete information related to number of cases, we should go with the approach involving least number of cases.

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