Bunuel wrote:
srini123 wrote:
Bunuel - the fourth approach you posted seems to have an error:
Fourth approach:
Total # of splitting group of 6 into two groups of 3: 6C3*3C3/2!=10
# of groups with Michael and Anthony: 1C1*1C1*4C1=4
P=4/10=40%
Total # of splitting group of 6 into two groups of 3: should be 6C3*3C3=20 and not 10.
Out of these 20 , there are 10 groups with Michael in it
so answer = groups with michael and anthony/ total groups with michael = 4/10 = 40%
Do we have group #1 and #2? Does it matter in which group Michael and Anthony will be? If no we should divide 6C3*3C3 by 2! and get 10 different split ups of 6 to groups of 3 when order of groups doesn't matter.
# of groups with Michael and Anthony together: 1C1*1C1*4C1=4
P=4/10=40%.
The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is not important is \(\frac{(mn)!}{(n!)^m*m!}\).
\(\frac{6!}{(3!)^2*2!}=10\)
OR another way:
In our case \(\frac{6C3*3C3}{2!}=10.\)
Sorry I may be missing something but, if we want to split 6 persons into 2 groups of 3 where order doesn't matter. dont we have 20 groups ?
for eg., let A,B,C,D,E,F be 6 people and the
groups with 3 people in each will be as below
ABC, ABD, ABE, ABF, ACD, ACE, ACF, ADE, ADF, AEF - 10
BCD, BCE,BCE, BDE,BDF, BEF - 6
CDE, CDE,CEF -3
DEF -1
total 20.
Let A be michael for this question , then total groups we have with michael is 10 and 4 groups have Michael and Anthony (assume D=anthony)
im sorry im still not getting why 20 is not the total groups irrespective of michael in it or not