The Interview

There are a few approaches to solving this problem.

• Method 1:

Choose one of the $3$ children to put last; then their guardian can be placed in one of the $5$ time slots before their child. In the next available time slot prior to that of the child in the last time slot place one of the remaining $2$ children; their guardian can then be placed in one of the $3$ remaining time slots before their child. The remaining two time slots can then be filled in only one appropriate way, (i.e., the remaining guardian prior to their child).

This gives us a total of $3*5*2*3 = \boxed{90}$ interview orderings.

• Method 2:

For any given guardian/child pairing, half of all interview orderings will have the guardian interviewed prior to the child. Since this is the case for each pairing, we have that $(\frac{1}{2})^{3} = \frac{1}{8}$ of all orderings will have each guardian come before their child. Since there are $6! = 720$ possible orderings, we have $(\frac{1}{8})*720 = \boxed{90}$ orderings that satisfy the given restriction.

• Method 3:

Label the guardians/children as $A, A, B, B, C, C$ . We don't actually have to distinguish between the guardians and their children in this labeling process, since when we permute these $6$ letters there is only one permitted way of ordering each guardian/child pairing. So the number of allowed orderings is the same as the number of permutations of these $6$ letters, which is $\dfrac{6!}{2! * 2! * 2!} = \dfrac{720}{8} = \boxed{90}$ .

one easy way is to consider a parent and a child pair and select 2 places for the pair. In this 2 places, they can be arranged in only 1 way. this can be done in C(6,2) ways. For the next pair it will be C(4,2) and for the third pair only 2 places are left, which leaves us at with only 1 way. so total possibilities= C(6,2)XC(4,2)XC(2,2)=15 6 1=90

I broke the problem into 3 cases.
Case 1: All 3 guardians come 1st and all the children come 2nd. 3!3! = 36 permutations by the fundamental counting principal.

Case 2: 2 guardians come 1st and the last guardian comes before their child. 3x2 ways to choose 2 guardians from 3. I listed the 6 ways one guardian can be placed in between the 3 children and yet come before their own. c1/g3/c2/c3, c1/c2/g3/c3, c1/g3/c3/c2, c2/g3/c1/c3, c2/c1/g3/c3, c2/g3/c3/c1 So total is 3x2x6=36

Case 3. 1 guardian comes 1st. There are 3 ways to do that. I listed the 6 ways to place the other two guardians among the 3 children and meet the condition. So total is 3x6=18

Total = case 1 +case 2 + case 3 = 36+36+18 = 90

Not elegant but that's how I did it.

There are 3 pairs of people. If we didnt care about the placement, this means there are 6! Or 720 ways they can be arranged. however we are told that each corresponding student guardian must be interviewed first which means that for each pair, half of the ways would have the guardian ahead ofnthe student. And becausebthere are thre pairs , its (1/2)^3 which equals to 1/8. therefore, the nber of ways in which the guardian is interviewed ahead of the student corresponding student is 1/8 of the total of number of ways or 90 ways.

Brute Force Solution:

• Must have child's guardian before that specific child, therefore the first person interviewed consist of the 3 possible guardians.
• From there you have your different scenarios which can be derived logically.
• In this case order matters so product rule is used, hence multiplication across a branch. Ex: There are 3 ways to pick the guardian "and" 2 ways to pick children ... etc.
• At the end, after completing each branch sum up all the possible ways and you get your solution.