Lets solve it without formula
Three condition is there
$1)$ Distinguishable job distinguishable employee
$2)$ Each employee got atleast 1 job
$3)$ Most difficult job to best employee
Say there are $J_{1},J_{2},J_{3}, J_{4},J_{5},J_{6},J_{7}$ and $4$ employees are $E_{1},E_{2},E_{3}, E_{4}$
Now say $J_{7}$ is most difficult job and $E_{4}$ is best employee
So, $J_{7}$ is fixed for $E_{4}$
Now, rest $6$ jobs we can divide in $4$ employees, So, that each employee can get $1$ job atleast
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First think about the combination $\left \{ 3,1,1,1 \right \}$
here 1,2,3.....,6 represents $J_{1},J_{2},J_{3}, J_{4},J_{5},J_{6}$ and only first $3$ ways of choosing we have shown ,because rest two we can choose only 1-1 ways
$\left \{ 1,5,6 \right \}$
$\left \{ 1,4,6 \right \}$
$\left \{ 1,3,6 \right \}$
$\left \{ 1,2,6 \right \}$
$\left \{ 1,4,5 \right \}$
$\left \{ 1,3,5 \right \}$
$\left \{ 1,2,5 \right \}$
$\left \{ 1,3,4 \right \}$
$\left \{ 1,2,4 \right \}$
$\left \{ 1,3,2 \right \}$
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$\left \{ 2,3,4 \right \}$
$\left \{ 2,3,5 \right \}$
$\left \{ 2,3,6 \right \}$
$\left \{ 2,4,5 \right \}$
$\left \{ 2,4,6 \right \}$
$\left \{ 2,5,6 \right \}$
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$\left \{ 3,4,5 \right \}$
$\left \{ 3,4,6 \right \}$
$\left \{ 3,5,6 \right \}$
$\left \{ 4,5,6 \right \}$
So, number of possible ways in $\left \{ 3,1,1,1 \right \}$ combination will be 20
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Now another combination possible here i.e. $\left \{ 2,2,1,1 \right \}$
here 1,2,3.....,6 represents $J_{1},J_{2},J_{3}, J_{4},J_{5},J_{6}$ and only first $2$ ways of choosing we have shown
Combinations are
$\left \{ 1,2 \right \}$
$\left \{ 1,3 \right \}$
$\left \{ 1,4 \right \}$
$\left \{ 1,5 \right \}$
$\left \{ 1,6 \right \}$
$\left \{ 2,3 \right \}$
$\left \{ 2,4 \right \}$
$\left \{ 2,5 \right \}$
$\left \{ 2,6 \right \}$
$\left \{ 3,4 \right \}$
$\left \{ 3,5 \right \}$
$\left \{ 3,6 \right \}$
$\left \{ 4,5 \right \}$
$\left \{ 4,6 \right \}$
$\left \{ 5,6 \right \}$
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Now for each first 2 combination, there is again $6$ combinations
here 1,2,3.....,6 represents $J_{1},J_{2},J_{3}, J_{4},J_{5},J_{6}$ and only second $2$ ways of choosing we have shown
Say, for $\left \{ 1,2 \right \}$ the $6$ combinations are
$\left \{ 3,4 \right \}$
$\left \{ 3,5 \right \}$
$\left \{ 3,6 \right \}$
$\left \{ 4,5 \right \}$
$\left \{ 4,6 \right \}$
$\left \{ 5,6 \right \}$
So, total number of ways will be $15\times 6=90$ i.e. 90
So, total number of ways 20+90=110
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If the same question I do by indistinguishable office and distinguishable employee, it will be answer as
$\binom{6+4-1}{4-1}=84$ ways
Is there anything I missing?