The number of partitioning of $\{1,\dots,10\}$ into $4$ groups such that no group has $1$ member

Question

Question :

How many ways can we partition $\{1,\dots,10\}$ into $4$ groups such that no group has just $1$ member?

Note 1 : This problem should be solved using inclusion-exclusion principle.

Note 2 : Should i consider the opposite case? What is it ? I mean, if "no group has just $1$ member" is false, then what is the other case? Should i consider that one time, one set has $1$ member , the other time, $2$ sets have $1$ member, the last time, $3$ sets have one member and the other one has $9$ members?

Answer 1

Every group must have at least $2$ members and Let the groups be $G1$ , $G2$ , $G3$ and $G4$

I can write as

$G1 + G2 + G3 + G4 = 10$ for $G(i) >= 0$,

But as the constraint is there that every group must have atleast 2 numbers, then I can modify the above equation as

$G1 + G2 + G3 + G4 = 2$ for every $G(i) >= 2$

This is a simple stars and bars problem with total number of ways = $C(5,2)$