A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed in
which at least one member of the group is a senior partner?
1 answer:
Answer:100
Step-by-step explanation:
There are ten total people:
10C3
10!/(7!×3!) = 120
Note: but that includes all the cases where there are no senior partners
So firstly let figure out the number of cases where there are no senior partners
because there are 6 junior partners
6C3
= 6!/(3!×3!) = 20
120 - 20 = 100 or the number of possible groups where there is at least 1 senior partner.
You might be interested in
I can't type it I have to solve on a paper for you
Answer:
5
Step-by-step explanation:
An image is translated and reflected.
Turn the 3/8 itto a mixed number and dived it by 111
A quotient of a number 21 is 3 and 63
Hope this helps!
