A wizard has $5$ friends. How many days does the wizards’ conference last? (Inclusion–Exclusion)

Question

A wizard has $5$ friends. During a long wizards’ conference, they met any given friend at dinner $10$ times, any given pair of friends $5$ times, any given threesome of friends $3$ times, any given foursome $2$ times, and all $5$ friends together once. The wizard ate alone $6$ times. Determine how many days the wizards’ conference lasted.

I think I'm supposed to use the Inclusion-Exclusion Formula.

Answer 1

This questions is really asking how many dinners occurred

Assume only 1 dinner per day

The problem gives us 6 pieces of information about how many friends the wizard ate dinner with. How many with 0 friends, how many with 1 friend, up to how many with all 5 friends.

The Inclusion-Exclusion Formula says

$$|A_1, A_2, A_3, ... A_{n-1}, A_n| = S_1 - S_2 + S_3 - S_4 + ... +(-1)^n S_{n-1} +(-1)^{n+1}S_n$$

Which means for a compound union of potentially overlapping sets $A_1, A_2, A_3, ... A_{n-1}, A_n$ that $S_1, S_2, S-3, S_4, ... , S_{n-1}, S_n$ are subsets and you add the subsets where an event is counted an odd number of times and subtract the subsets where an event was counted an even number of times. The giant formula turns into "add $S_1, S_3, S_5, ...$ and subtract $S_2, S_4, S_6, ...$"

The 6 pieces of information from the problem correspond to these subsets

$S_1={5 \choose 1}*10$ We have to choose 1 of the 5 friends ("it met any given friend at dinner 10 times")
$S_2={5 \choose 2}*5$ How many ways to choose 2 friends ("any given pair of friends 5 times")
$S_3={5 \choose 3}*3$ How many ways to choose 3 friends ("any given threesome of friends 3 times")
$S_4={5 \choose 4}*2$ How many ways to choose 4 friends ("any given foursome 2 times")
$S_5={5 \choose 5}*1$ How many ways to choose 5 friends ("all 5 friends together once")

$S_1 - S_2 + S_3 - S_4 + S_5$ Is how many nights the wizard ate with at least 1 friend
But we can't forget about the 6 times he ate alone

The total number of dinners is
$S_1 - S_2 + S_3 - S_4 + S_5 +6$
$={5 \choose 1}*10 - {5 \choose 2}*5 + {5 \choose 3}*3 - {5 \choose 4}*2 + {5 \choose 5}*1 + 6$
$=5*10 - 10*5 + 10*3 - 5*2 + 1*1 + 6 $
$=50 -50 +30 -10 +1 +6$
$=27$

Answer 2

For convenience I'll call the wizard of the title Rincewind ($R$) who has friends Antares ($A$), Betelgeuse ($B$), Castor ($C$), Deneb ($D$) and Enif ($E$). I'm assuming here that a dinner in a larger grouping also counts towards the smaller contained groupings.

The required dinners are, per grouping level, $(10,5,3,2,1)$, plus $6$ lonesome dinners for $R$.

So one dinner is $RABCDE$. This takes care of the "all-five-friends" dinner and one of each of the other categories, so we have $(9,4,2,1,0)$ left to allocate.

Then we have five dinners $RBCDE$, $RACDE$, $RABDE$, $RABCE$, $RABCD$ to satisfy the sets of four. This also caters for a number of other required sets, leaving $(5,1,0,0,0)$.

So then $\binom 52 = 10$ dinners for two friends, $RAB$, $RAC$, $RAD$, $RAE$, $RBC$, $RBD$, $RBE$, $RCD$, $RCE$, $RDE$. This is leaves just $(1,0,0,0,0)$ to fulfill.

So finally five intimate dinners with each one of $R$'s friends, $RA$, $RB$, $RC$, $RD$, $RE$.

Total: $6+1+5+10+5 = 27$ dinners.

Wizards being unstoppable gluttons, they may well have had three dinners a day, which would give a nine-day conference.