Question 718179: Number of ways in which 8 students can be divided into 2 teams , that need not necessarily be of equal size.
Ans 127
My method
As 8 student is divided into 2 teams
Team A==============Team B
1 =============== 7 students===== 8C1*7C7 - equation(1)
2 =============== 6 students ===== 8C2 -(2)
3 =============== 5 students ===== 8C3 -(3)
4 =============== 4 students ===== 8C4 -(4)
5 =============== 3 students ===== 8C5 -(5)
6 =============== 2 students ===== 8C6 -(6)
7 =============== 1 student ===== 8C7 -(7)
Therefore no. of ways = (1)+(2)+(3)+(4)+(5)+(6)+(7)=252
But actual answer is the half of this answer.Can you tell me where I go wrong?
Answer by oscargut(2103)   (Show Source):
You can put this solution on YOUR website! Team A
1 student: C(8,1) = 8
2 students: C(8,2) = 28
3 students: C(8,3) = 56
4 students: C(8,4) = 70
5 student: C(8,5) = 56
6 students: C(8,6) = 28
7 students: C(8,7) = 8
Answer: 254
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