agarwalaayush2007
Bunuel
QZ
In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
A. 159
B. 208
C. 209
D. 212
E. 215
(At least 1 boy) = (Total) - (0 boys, so all girls)
(At least 1 boy) = 10C4 - 4C4 = 10!/(6!4!) - 1 = 209.
Answer: C.
Thank you for the explanation Bunuel!

However, could you please explain what is wrong in the below approach:
atleast one guy - 6C1 multiplied by 9C3 - ways to select three people from the remaining 9.
The above approach gives a wrong answer but i am unable to identify the error in logic. Please help!
This way will give you duplication.
Say, the boys are {1, 2, 3, 4, 5, 6} and girls are {a, b, c, d}.
6C1 gives you one boy from six, say {1}.
9C3 gives you three children from nine, say {2, 3, a}.
So, the group is {1, 2, 3, a}.
But we will get the same group in other ways too. For example:
6C1 can give you one boy from six, say {2}.
9C3 can give you three children from nine, say {1, 3, a}.
So, the group is {1, 2, 3, a}.
As you can see we got the same group of children.