Determine the number of ways in which 8 people can be divided into 2 equal groups.
Hint: You need to pick 4 people from 8. Combination problem.
dont you just use combinations? \[nCr = \frac{n!}{r!(n-r)!}\] \[8C2 = \frac{8!}{2!(8-2)!}\]
uh why do you need to pick 4 people from 8?
2 Equal groups. How many people will be there in one group 4 right?
yeh
Now you need to select 4 out of 8.
but 8 people right? and we want two groups? so 4 people in one group but you only pick one or...?
so confusing....
Nope. How many people you need to pick? Equal groups means 4 +4. So you will pick 4 out of eight to make a group. In 8C4 ways you can pick 4 people from 8 right?. What's confusing?
It doesn't say how many people to pick or anything just basically how many groups we can divide evenly?
oh k, I think I got it now, don't worry.
Okay let's say A B C D E F G H are 8 persons. How many ways you can divide equally? ( A B C D) ( E F G H) + ( A B C E) (D F G H)+ and so on right? Unknowingly you are picking right
yeh.
Mathematically we will write 8C4 right?
yep.
I get it now, but can you help me with this last example.
The number of ways in which 10 objects can be divided into 2 unequal groups is: A 385 B 835 C 950 D 640 E 565
I think it's something - 10C5
tough one
10C1 + 10C2 + ........ 10C10 - 10C5 assuming a group can have 0 objects
each term represents number of ways to chose objects for first group
but there's no answer for that.
the answer for the way u did it @rsadhvika is 771 and there's no option for that.
2(1+10+45+120+210) it cannot be a odd number
772
ok its not there in the options either
ok well it's late over here so I got to sleep now, if you can get the answer just show me the working, thanks.
I'm litle bc. I'll try this later..:)
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