Acertain 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?

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.

(-5,-28)

i graphed to on desmos.