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As this is a question of forming a Team, hence we apply Combination and Not Permutation.

Total no. of ways to form teams of 3 members
= 1 Sr.Part.* 2 Jr.Part. + 2 Sr.Part.* 1 Jr.Part. + All 3 Sr.Part.* No Jr.Part.

=> 4C1*6C2 + 4C2*6C1 + 4C3

=> 4*6.5/1.2 + 4.3/1.2*6 + 4 ; As 4C3 = 4C1.

=> 100. Ans B.

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Shalabh Jain,
e-GMAT Instructor

1 Sr 2 Jr + 2 Sr 1 Jr + 3Sr 0Jr
4C1 * 6C2 + 4C2 * 6 + 4C3
4*15 + 36 + 4
100

IMO B.

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Getting defeated is just a temporary notion, giving it up is what makes it permanent.
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10C3- 6C3= 100

Its B

Can you elaborate it a bit more ? Will be helpful for everyone as well.

_________________
Getting defeated is just a temporary notion, giving it up is what makes it permanent.
http://gmatandbeyond.blogspot.in/

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? (Two groups are considered different if at least one group member is different.)

total number of groups possible = (4+6)C3 = 10C3= 120

Required number of groupings possible
= total groupings - groupings with no senior partner
= 120- 6C3
= 120- 20
= 100

Hope this helps..!!