Factorial Word Problem

Question

Here is the problem:

"The traditional French greeting is to kiss the other person on each cheek. So, when two traditional French friends meet, one kisses the second on both cheeks and the second person kisses the first on both cheeks. a. Make an In-Out table where the In is the number of friends meeting and the Out is the total number of kisses. Your Ins should go from 1 through 6 at least. b. Write a formula for the total number of kisses, k, in terms of the number of friends, f."

So I think that the formula has something to do with factorial but I cannot find a formula. I have a table with points (1,2), (2,6), (3,24), (4,120), (5,720) and (6, 5040). Where x is the number of friends you meet with and y is the total number of kisses. Is this correct?

Answer 1

The counts do not seem to be right. If there is only $1$ person in the group, there are $0$ kisses.

If there are $2$ people in the room, there are $4$ kisses (A kisses B on both cheeks, $2$ kisses, B reciprocates, $2$ more).

Now look at the case of $3$ people, A, B, and C. Again we can make an explicit count. For A and B, we have $4$ kisses. For A and C, $4$ more, and finally for B and C, $4$ more, for a total of $12$.

If you have a certain amount of patience, you can deal with $4$ people, A, B, C, and D. First A exchanges kisses with B, C, and D. That's $12$ kisses. Now B, C, and D exchange kisses with each other. That's the case of $3$ people, which we have already solved! So they account for $12$ more, for a total of $24$.

Now deal with $5$ people, A, B, C, D, and E. First A does the kiss thing with B, C, D, and E. That's $16$ kisses. Then B, C, D, and E exchange kisses. That's already been computed, it is the $4$ people case, we get $24$. so the total is $16+24$. which is $40$.

Finally, deal with $6$ people, A to F. First A exchanges kisses with the $5$ others, total $20$. Then the remaining $5$ exchange kisses, total $40$. The sum is $60$.

To get a general formula, there are many approaches. We could follow up on our analysis above. Or we could start again.

For every pair of friends, there will be $4$ kisses. Do you know a simple formula for the number of pairs there are if there are $f$ friends in the group?

If you do not, here is a way of counting. We will call the people by the even more boring names $1,2,3,4,\dots,f$. Let us first count the number of ordered pairs $(x,y)$ that we can form. The first person in the ordered pair can be chosen in $f$ ways. For each of these ways, the second person can be chosen in $f-1$ ways. So the total number of ordered pairs is $f(f-1))$. But this is not the number of plain pairs, for it counts (Alicia, Beti) as different from (Beti, Alicia). The fix is simple: divide our count of ordered pairs by $2$. So the number of pairs of people is $$\frac{f(f-1)}{2}.$$

Now make your formula. To check for any mistakes, compare what your formula predicts with our explicit counts above.

Answer 2

It is not quite related to the factorial but to the number of pairs you can form with the friends. If there are $N$ friends then how many unique pairs of friends are there? Each pair will exchange $4$ kisses so your total number of kisses is $4$ times the number of pairs.

Edit: Some elaboration.

Consider the choose function $\binom{n}{r}$. If we have $f$ friends and we need to find out the number of pairs, then that's equivalent to choosing $2$ out of the $f$ friends. Therefore the number of pairs is given by $\binom{f}{2}$. The binomial coefficients are normally given by factorials, but in this case it simplifies drastically. $$\binom{f}{2} = \frac{f!}{(f-2)!2!} = \frac{f(f-1)}{2}$$ The number of kisses is $4$ times this value, so the total number of kisses is given by $$k = 2f(f-1)$$