In how many ways can we arrange $7$ different things to $3$ people, such that all of them must get at least one?
We know that if we have $n$ identical items which will be distributed in $r$ distinct groups where each must get at least one then the number of way is $\binom {n-1}{r-1}$ (i.e in this case $\binom62$. But in the question it is said that $7$ DIFFERENT things then what will be the approach?
MY TRY ::
At first selecting $1$ by $1$ for each $3$ people so that each get at least one : $\binom71 \times \binom 61 \times \binom 51$
Now each of $4$ can go to any of $3$ so :$3^4$ ways, so $\binom71\times\binom61\times\binom51\times (3^4)$
MY TRY #2 ::
total possibilities – {any one get $0$ $(0,6,1)(0,5,2)(0,4,3)$} – {any two get $0$ $ (0,0,7)(0,7,0)(7,0,0)$}
so
$3^7$ – {($\binom76 + \binom 75 + \binom 74$ )$\times$ 3!} – 3
Best Answer
It's actually helpful here to generalize the problem to distributing $n$ different objects to $3$ people (with $n\ge3$ so that each person can be guaranteed getting at least one item). If you disregard the requirement that each person get at least one item, then you get the overestimate $3^n$. From this, let's subtract the number of ways you can pick one person to not get any items, distributing the rest to the other two, giving
$$3^n-3\cdot2^n$$
But this now under-estimates because you're now doublecounting occasions when two people fail to receive any items. So we need to add back in the number of ways this can happen, producing
$$3^n-3\cdot2^n+3\cdot1^n$$
As a sanity check, consider the case $n=3$, where there are clearly $6$ ways to assign an item to each person:
$$3^3-3\cdot2^3+3=27-24+3=6$$
For $n=7$, the formula gives
$$3^7-3\cdot2^7+3 = 2187-3\cdot128+3=1806$$
The general principle here is known as inclusion-exclusion. Note that it gives the correct answer, $0$, even for $n=1$ and $2$.