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If Jack/Jill is not selected as D, then D has 3 choices, and P has 5 choices.
Therefore, no. of choices = 3*5

But is this not possible in 2 different ways? Like, if Jack/Jill is selected as D, then we have 3*5 choices. But the other way - if Jack/Jill is selected as P, we again have 3*5 choices. Right??
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kiran120680
A company interviewed 5 applicants each for the posts of the Director and the President. If Jack and Jill were the only applicants who were interviewed for both the posts and an applicant can be selected for only one of the posts, what is the number of ways in which the company can select its Director and President from the interviewed applicants?

A. 20
B. 23
C. 25
D. 27
E. 35

Simplification of the question can be .... There are 2 common candidates in different sets. So unique candidates in each group are 3

D P
3 * 3 = 9
Jack *3 = 3
Jill *3 = 3
3 * Jack = 3
3 * Jill = 3
Jack * Jill = 1
Jill * Jack =1

Total=23

Hope ...I'm correct!...
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kiran120680
A company interviewed 5 applicants each for the posts of the Director and the President. If Jack and Jill were the only applicants who were interviewed for both the posts and an applicant can be selected for only one of the posts, what is the number of ways in which the company can select its Director and President from the interviewed applicants?

A. 20
B. 23
C. 25
D. 27
E. 35

Jack = X, Jill = Y

CASES:
X for D and Y for P: 1*1=1
Y for D and X for P: 1*1=1
X for D and ≠Y for P: 1*3C1=3
X for P and ≠Y for D: 1*3C1=3
Y for D and ≠X for P: 1*3C1=3
Y for P and ≠X for D: 1*3C1=3
Not X or Y for any: 3C1*3C1=9

Total: 2+3(4)+9=2+12+9=23

Ans (B)
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kiran120680
A company interviewed 5 applicants each for the posts of the Director and the President. If Jack and Jill were the only applicants who were interviewed for both the posts and an applicant can be selected for only one of the posts, what is the number of ways in which the company can select its Director and President from the interviewed applicants?

A. 20
B. 23
C. 25
D. 27
E. 35

If both Jack and Jill are selected, then there are 2 ways: Jack is the Director and Jill is the President, and vice versa.

If Jack is selected and Jill is not, then there are 1 x 3 = 3 ways if Jack is selected as the Director and another 3 ways if he is selected as the President. Therefore, there are a total of 6 ways if Jack is selected and Jill is not.

Likewise, there are a total of 6 ways if Jill is selected (as either the Director or the President) and Jill is not (selected at all).

Finally, if neither Jack nor Jill is selected, then there are 3 x 3 = 9 ways for the Director and the President to be chosen.

Therefore, there are a total of 2 + 6 + 6 + 9 = 23 ways.

Alternate Solution:

If Jack is selected as the Director, then he cannot be selected as the President, and thus, there are 4 choices (one of which is Jill) for the position of President. Thus, the company can fill the positions in 4 ways when Jack is the Director.

If Jill is selected as the Director, then she cannot be selected as the President, and thus, there are 4 choices (one of which is Jack) for the position of President. Thus, the company can fill the positions in 4 ways where Jill is the Director.

Suppose neither Jack nor Jill is selected for the position of Director. This means that one of the 5 - 2 = 3 candidates is chosen for this position. Further, since neither Jack nor Jill is selected as the Director, there are 5 candidates for the position of President (including Jack and Jill). Thus, there are 3 x 5 = 15 ways the company can fill the positions where neither Jack nor Jill is the Director.

In total, there are 4 + 4 + 15 = 23 ways the company can fill the positions.

Answer: B
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Its simple there are 5x5 ways of selecting candidates
Now out of 25 ways there are only two ways which are not possible - one is jack at both places & other is Jill at both places
therefore total posiible ways = 25-1-1=23
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aashraiarun
If Jack/Jill is not selected as D, then D has 3 choices, and P has 5 choices.
Therefore, no. of choices = 3*5

But is this not possible in 2 different ways? Like, if Jack/Jill is selected as D, then we have 3*5 choices. But the other way - if Jack/Jill is selected as P, we again have 3*5 choices. Right??
­No because he makes a case of having this restriction only in D let's say and he continues with forcing Jack/Jill in D. What you said double counts the 3*3 and in the next step you will double count more things
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