Suppose you are managing 19 employees, and you need to form three teams to work on different projects. Assume that all employees will work on a team, and that each employee has the same qualifications/skills so that everyone has the same probability of getting choosen. In how many different ways can the teams be chosen so that the number of employees on each project are as follows:

6 2 11

First case: 325909584 is number when teams have at 6 employees.

Second case: 2441880 is number when teams have at 2 employees.

Third case: This is not possible because three teams with 11 employees cannot be created, because the total number of employees is 19.

Step-by-step explanation:

We have managing 19 employees, and you need to form three teams to work on different projects.

First case: in team have 6 employees.

We calculate number of way, and we get

/frac{19·18·17·16·15·14·13!}{ (19-6) ! · 6!}=19·17·2·3·14 = 27132

/frac{13·12·11·10·9·8·7!}{ (13-6) ! · 6!}=13·11·2·3·2 = 1716

/frac{7·6!}{ (7-1) !· 1!} = 7

so, we have:

27132·1716·7 = 325909584 is number when teams have at 6 employees.

Second case: in team have 2 employees.

We calculate number of way, and we get

/frac{19·18·17!}{ (19-2) ! · 2!}=19·9 = 171

/frac{17·16·15!}{ (17-2) ! · 2!}=17·8 = 136

/frac{15·14·13!}{ (15-2) !· 2!} = 15·7=105

so, we have:

171·136·105 = 2441880 is number when teams have at 2 employees.

Third case: in team have 11 employees.

This is not possible because three teams with 11 employees cannot be created, because the total number of employees is 19.