7 jobs to be given to 4 employees and each one must get at least 1 job. But the most difficult one goes to the best employee. So, the problem reduces to
- Giving 6 jobs to 3 employees such that each one gets 1 job (here best employee gets exactly 1 job which is the toughest)
- Giving 6 jobs to 4 employees such that each one gets 1 job (here best employee gets at least 2 jobs).
Now, 1 is same as putting 6 identical balls to 3 distinct bins such that no bin is empty and is given by $3! \times S(6,3)=90*6 = 540$ and similarly 2 is given by $4!*S(6,4)=65*24 = 1560$. So, total ways = $540+1560=2100$. $S(m,n)$ is Stirlings number of second kind and can be derived using the triangle given below,
$S(m,n) = S(m-1,n-1) + nS(m-1, n), m,n > 1$
$S(m,1) = S(m,m) = 1$
1
1 1
1 3 1
1 7 6 1
1 15 25 10 1
1 31 90 65 15 1