Armstead Productions conducted two rounds of interviews for

Question

Armstead Productions conducted two rounds of interviews for 5 candidates. In each round, each candidate was assigned at random to one of the company’s 5 managers so that each manager interviewed exactly one candidate.

What is the probability that exactly 3 of the 5 managers interviewed more than one of the 5 candidates ?

(A)  \frac{1}{12} 
(B)  \frac{1}{10} 
(C)  \frac{1}{8} 
(D)  \frac{2}{15} 
(E)  \frac{1}{6}

ankushsambare · Answer

Think of the first round as fixed.
In the second round, each of the 5 candidates is again assigned to the 5 managers, one candidate per manager. Thus the second round corresponds to a permutation of the 5 candidates.
A manager interviews the same candidate in both rounds iff that manager is a  fixed point  of the permutation.
The question asks for the probability that  exactly 3 managers interviewed more than one candidate .
Since each manager interviews exactly one candidate in each round:
 
 If a manager gets the same candidate both times, they interviewed only 1 distinct candidate. 
 If a manager gets a different candidate in round 2, they interviewed 2 distinct candidates.  
Therefore, "interviewed more than one candidate" means  not a fixed point .
Exactly 3 managers interviewed more than one candidate
⇒ exactly 2 managers interviewed the same candidate both rounds
⇒ the permutation has exactly 2 fixed points.
Total permutations of 5 candidates:
5! = 120
Count permutations with exactly 2 fixed points:
 
 Choose the 2 fixed points:
C(5,2) = 10 
 The remaining 3 candidates must have  no fixed points  (a derangement of 3).
Number of derangements of 3:
!3 = 2  
Thus favorable permutations:
10 × 2 = 20
Probability:
20 / 120 = 1/6
Answer:  E. 1/6.

BongBideshini · Answer

Suppose in 1st round
 
 M1 interviewed C1 
 M2 interviewed C2 
 M3 interviewed C3 
 M4 interviewed C4 
 M5 interviewed C5  
Now look only at Round 2.
A manager interviewed  more than one candidate  if the candidate in Round 2 is different from the candidate in Round 1.
So:
 
 If M1 gets C1 again, M1 interviewed only one candidate total. 
 If M1 gets C3, M1 interviewed two candidates (C1 and C3).  
We need exactly 3 managers to get a different candidate in Round 2; 2 managers will get the same candidate as round one.

Choose the 2 fixed managers who get the same candidates as in the 1st round.
C(5,2) = 10
The remaining 3 managers must have no candidates. The number of derangements of 3 elements is:
!3 = 2
Therefore, favorable permutations:
10 × 2 = 20

Total permutations:
5! = 120

Probability:
20/120 = 1/6
 Answer: E

Note:  
 Derangement formula: 
 !n = n! × (1 − 1/1! + 1/2! − 1/3! + ... + (−1)^n / n!) 
where !n means the number of ways to arrange n objects so that no object remains in its original position.
!1 = 0
!2 = 1
!3 = 2
!4 = 9
!5 = 44

Regor60 · Answer

First determine the base against which the successful combinations will be measured.

Here we will use permutations so the base is how many ways can five candidates be assigned to five interviewers:

5! = 120

Exactly three interviewing more than one candidate means exactly two interview the same candidate twice.

Number of ways those two candidates can be selected:

5!/2!3! = 10

Now, since we're using permutations, those two can be arranged in only one way.

The remaining three have to be assigned to new interviewers for the next round with no duplication. Since the numbers are relatively small we will assign interviewee and interviewer as follows A1, B2, c3, D4, and E5.

If we assign A1 and B2 as those repeating the interviewer then we can see that C can be assigned to D or E and that in each case D and E in turn can only be assigned one way. So two ways to assign the remaining three the different interviewers.

So total ways is 10*2 = 20

And therefore the probability is 20/120 = 1/6

Regor60 · Answer

Another way using probability.

Again, three interviewers interviewing more than one candidate means exactly two interviewers interview the same candidate twice.

There are 5!/3!2! = 10 ways to select.

The probability that one of the two candidates is assigned their original interviewer is 1/5. Therefore the probability that the second of the two candidates is assigned their original interviewer is now 1/4.

The probability that the third candidate is assigned a different interviewer is 2/3. Since there are now two interviewers left for the remaining two interviewees but only one of those ways creates a new interviewer interviewee pair that probability is 1/2.
Therefore the total probability is 10*1/5*1/4*2/3*1/2 =10*1/60 = 1/6

Armstead Productions conducted two rounds of interviews for

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