This question has been solved but I have a specific question that has NOT been addressed. My exam is next week - will appreciate expedited response.
A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formd in which at least one member of the group is a senior partner? (Two groups are considered different if at least one group member is different)
A) 48
b) 100
c) 120
d) 288
e) 600
When I looked at this question, I wanted to apply the SLOT method. And this is how I thought about it
3 slots: we must have one senior partner, and the other 2 slots can be filled by any body
4*9*8=288
Now, since order does NOT matter -- we're picking "groups" of 3 here...we must divide by the factorial of the number of inter-changeable items.
There are 3 inter-changeable items.
Let senior partner = SP and junior partner = JP
SP, JP, JP
JP, SP, JP
JP, JP, SP
*are all the same
Thus, we divide 288/3! = 48
Note: my interpretation of a group is that it doesn't matter how the 3 items are arranged within that group. And it doesn't matter what those 3 items are.
The OA is 100.
HOW and WHY is my approach incorrect?