If 3 members are to be selected from a team of x members then what is

Isnt xC3-(x-1)C3=15??

7C3= 35 and 6C3 =20, or am I wrong?

Actually, it shouldnt be (x-1)C3.. The no of ways of selecting 3 out of x, so that two specific members are together on the team = (x-2)C1.. this is because if we have already fixed two members, then we have to select remaining 1 more member out of (x-2).

So it should be xC3 - (x-1)C2 = 30
And you can check: 7C3 - 5C1 = 35-5 = 30

Bunuel,

Is (x-2)C1 correct in chetan2u answer ?

The total way of selecting a team of r members out of n members is nCr.
Thus, the total ways of selecting a team of 3 members out of x members will be xC3.

1. If the team of 3 members is selected from the group of x members, then 30 different combinations of 3 members can be formed with 2 specific members never being together on the team.

Total ways of selecting 3 members from the group of x members with 2 specific members never being together on the team = (Total ways of selecting 3 members from the group of x members) - (Total ways of selecting 3 members from the group of x members with 2 specific members always being together on the team)

Total ways of selecting 3 members from the group of x members = xC3

Total ways of selecting 3 members from the group of x members with 2 specific members always being together on the team = x (since only one member is left to pick)

Thus, 30 = xC3 - x = (x)(x-1)(x-2)/6
30 x 6 = x(x-1)(x-2)
Or x = 7
Thus, this is sufficient.


2. There are a total of 35 ways to make a team of 3 members out of the team of x members.


Total ways of selecting 3 members from the group of x members = xC3

Thus, 35 = xC3
35 = x(x-1)(x-2)
Or, x = 7
Thus, this is sufficient.

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