A certain company consists of 3 managers and 8 non-managers. How many different teams of 3 employees can be formed in which at least one member of the team is a manager and at least one member of the team is not a manager? (Two groups are considered different if at least one group member is different)

A. 84
B. 108
C. 135
D. 270
E. 990

I solved this task in this way:

(11!/3!*8!)-(3!+8!/3!5!)=103,
where (11!/3!*8!) is the total number of combinations and (3!+8!/3!5!) - undesirable output, e.g. only managers or non managers are in the teams
Could please someone to help me to point out, where exactly my logic comes wrong?

i got it: instead of 3! i had to use 3!/3! for the combination of managers in a team
Bunuel! as always - thank you!

hi guys, first post
i have a problem in understanding this question and why my logic is wrong, this is how i try to solve the problem:

i use the formula n!/(n-k)!*k!

so first n=3 and k=1 because we have 3 managers to choose from and can choose one
3!/(2!*1!)
second n=3 and k=2 because we have 3 managers to choose from and can choose two
3!/(2!*1!)
third n=8 and k=1 because we have 8 non-managers to choose from and can choose one
8!/(7!*1!)
fourth n=8 and k=2 because we have 8 non-managers to choose from and can choose two
8!/(6!*2!)
then i multiply all the parts= 3*3*8*28=2016
can someone please help me
thank you

Selection of a Manager from 3 Managers AND
Selection of 2 Non-Managers from 8 Non-Managers OR
Selection of 2 Managers from 3 Managers AND
Selection of a Non-Manager from 8 Non-Managers

AND = Multiplication, OR = Addition

3C1 * 8C2 + 3C2 * 8C1 = 3 * 28 + 3 * 8 = 24 + 84 = 108
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The following post might help: a-five-member-committee-is-to-be-formed-from-a-group-of-five-55410.html#p1248236
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Dear Bunuel

I do not understand the formulas here:

C13∗C28+C23∗C18=84+24=108. > How do you calculate 84 and 24 ... it must be with factorials but I do not get the way you write it. thank you

Hi reto

C13 is just another way of writing 3C1.

So the expression here can be converted to

3C1*8C2 + 3C2*8C1
= 3*(8*7/2) + 3*8
= 84 + 24
=108