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In how many ways can a group of six teachers break into four groups, so as to guard four examination rooms.

Given answer : 1560.

Let the teachers be: $A, B, C, D, E, F.$ and the rooms be: $1, 2, 3,4.$

Must need at least one teacher in a room.
So, let the groups be $G_1, G_2, G_3, G_4.$
And, there can be any one of the six teachers be assigned to $G_1.$ And out of the rest, can assign one teacher to $G_2$ in five ways. Continuing, in this way, have : $ 6\cdot 5\cdot 4\cdot 3= 360$ ways.

The rest of the two teachers can be assigned to any of the four groups in $4 P 2= 12$ ways

Multiplying together get: $360\cdot12= 4320$ ways.
This differs from the stated answer of $1560?$

N. F. Taussig
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jiten
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    Your method overcounts badly. For instance, you count $ABC,D, E, F$ three times. Once taking $A$ to be the teacher initially placed in $G_1$, then taking $B$, then taking $C$. Hard to repair the method. Better off with Inclusion-Exclusion. Or, since the numbers involved are so small, simply write out the possible patterns and count the ways to populate each of them. – lulu Oct 06 '22 at 10:39
  • @lulu Am not clear how am counting a given composition of the four groups : $ABC, D, E, F,$ thrice? – jiten Oct 06 '22 at 10:50
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    I explained that. You start by choosing one teacher for each room, then assign the others. So if you start with $A,D, E,F$ and assign $B, C$ to room $1$ then you get $ABC, D,E,F$. But if you start with $B,D,E,F$ and assign $A,C$ to room $1$ then you also get $ABC,D,E,F$. So you have counted the same assignment twice. Or you could start with $C,D,E,F$ and assign $A,B$ to group $1$. That's three times. – lulu Oct 06 '22 at 10:52
  • @lulu Repair seems uncertain, as need to subtract such cases; and have no idea. Inclusion-Exclusion works by taking all possible cases, and then excluding those counted multiple times. But, that reduces to finding possible patterns. The patterns can be in the allocation of the left two teachers to any group. These are: $1,0,0,1; 1,0,1,0; 1,1,0,0; 0,0,1,1; 1+1,0,0,0; \cdots$ But, cannot see how that helps. Better if some more hints be given. – jiten Oct 06 '22 at 11:03
  • I proposed two methods to solve the problem. Inclusion-Exclusion is routine, and not particularly difficult. The second method, enumerate the patterns and count each type separately, is probably easier in this case but that's only because you have such small numbers. In general, there will be lots of patterns which you'll need to list and analyze. But in this case it's easy. – lulu Oct 06 '22 at 11:05
  • @lulu Sorry, need some hints as say, for IE approach, am not clear what to exclude. Seems another form of finding patterns. For finding patterns, need fix a particular selection of teachers as: $A, D, E, F.$ Then, $ABC, D, E,F; A, DBC, E, F; A, D, EBC, F; A, D, E, FBC; AB, CD, E, F; AB, C, ED,... $ Sorry, am lost. – jiten Oct 06 '22 at 11:20
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    This is a routine problem. Assign the teachers without constraint, that's $4^6$ assignments. Unfortunately, you might have left a room (or rooms) empty. So now exclude the case where you have an empty room. Then restore the ones with two empty rooms, and so on. – lulu Oct 06 '22 at 11:26
  • @lulu The universe is the possible assignment of each teacher to anybody the four rooms, leading th $4^6$ ways. Then, for one empty room case have $3^6$ choices, and as are counting permutations; so multiply by four, to get: $729\cdot 4= 2916.$ Next, need consider for two empty rooms: $2^6$ choices, but need exclude the cases covered under that for the case of one empty room. Then need multiply that by $4P2= 12.$ But, not clear how to perform the subtraction. Once able to find that, can take the left case for three empty rooms. – jiten Oct 06 '22 at 11:46

2 Answers2

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This may not be how you want to do it, but it's how I usually count the surjections of an $m$-set on to an $n$-set.

Step 1. Split the $6$ teachers into $4$ non-empty parts. This must be into (a) $3+1+1+1$ or into (b) $2+2+1+1$. Case (a) occurs in $({6\choose 3}{ 3\choose 1} {2\choose 1} {1\choose 1})/ 1! 3!=20$ ways. Case (b) occurs in ${6\choose 2}{4\choose 2}{2\choose 1}{1\choose 1}/2!2!=45$ ways.

Step 2. Assign the $4$ sets of teachers to classrooms. That can be done in $4!=24$ ways.

Step3. The total number of ways is therefore $24(20+45)=1560.$

ancient mathematician
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This is an Inclusion-Exclusion problem.

See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.

First, I will give the corresponding enumeration. Then I will explain it.

The enumeration is

$$4^6 - \left[\binom{4}{1}3^6\right] + \left[\binom{4}{2}2^6\right] - \left[\binom{4}{3}1^6\right] + \left[\binom{4}{4}0^6\right].$$

Let $S$ denote the set of all of the ways that the teachers can be assigned, without any regard to forcing at least one teacher to be assigned to each room.

For $k \in \{1,2,3,4\}$ let $S_k$ denote the subset of $S$ where room $(k)$ has no teacher assigned.

Then, the desired enumeration is

$$|S| - |S_1 \cup S_2 \cup S_3 \cup S_4|.$$

Let $T_0$ denote $|S| \implies T_0 = 4^6.$

That is, there are $(4)$ choices for each teacher.

For $r \in \{1,2,3,4\}$, let $T_r$ denote the $\binom{4}{r}$ terms represented by

$$\sum_{1 \leq i_1 < i_2 < \cdots < i_r \leq 4} ~\left|S_{i_1} \cap \cdots \cap S_{i_r}\right|. \tag1 $$

By considerations of symmetry, each term in the computation of $T_r$ will be equivalent to the computation of

$$|S_1 \cap \cdots \cap S_r|. \tag2 $$

In (2) above, such a computation represents that there are $(4-r)$ choices for each teacher.

Therefore,

  • $\displaystyle T_r = \binom{4}{r} (4-r)^6.$

  • Per Inclusion-Exclusion theory, the desired computation is

    $\displaystyle \sum_{r=0}^4 (-1)^r T_r.$

user2661923
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  • $4^6 - (3^6)\cdot 4 + 2^6\cdot 6 - 4 +1= 4096 -729\cdot 4 + 64\cdot 6 -3 = 4096 - 2916 + 384 -3= 1561.$ That shows the case of all empty rooms is not to be counted. – jiten Oct 06 '22 at 12:30
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    @jiten You are wrong. The case of all empty rooms is to be counted. It is just that you miscalculated $(0)^6.$ That is, $T_4$, which is to be counted, evaluates to $(0)$. – user2661923 Oct 06 '22 at 12:32