HeroicPandas
Heroic!
- Joined
- Mar 8, 2012
- Messages
- 1,546
- Gender
- Male
- HSC
- 2013
1) Eight people are to be divided into 2 groups. What is the probability that there will be 4 in each group.
My go
 = $\frac{\binom{8}{4}\binom{4}{4}}{\binom{8}{1}\binom{7}{7} + \binom{8}{2}\binom{6}{6} + \dots + \binom{8}{7}\binom{1}{1}} = \frac{\binom{8}{4}\binom{4}{4}}{\sum_{k=1}^{7} \binom{8}{k}} = \frac{35}{127} \neq \frac{35}{81})
NOTE:Some LaTeX error: 'm77' is 7C7
2) 6 humans, of whom A and B are two, arrange themselves at random in a row. What is the probability that there are at least 3 humans between A and B?
my go
 =P($ 3 between) + P(4 between) \\ \\ Prob = $\frac{(2!)(2)\binom{4}{3}(3!)}{6!} + \frac{(2!)\binom{4}{4}(4!)}{6!} = \frac{1}{5} \neq \frac{2}{5})
For question 2, my explanation is:
For 3 between:
[A _ _ _ B] _
Deal with elements between A and B. A and B are gone, 4 are left, so choose 3 out of 4 (4C3) then times by 3! for their arrangements
Arrangements of individual elements (the grouped and single one) = 2!
A and B can switch places = 2
Therefore = 2 x 2! x 4C3 x 3!
For 4 between:
[A _ _ _ _ B]
Elements between A and B = 4C4 x 4!
A and B can switch = 2
Therefore = 2 x 4! x 4C4
what am i doing wrong??
I'd appreciate help with this
(I am not good at this)
My go
NOTE:Some LaTeX error: 'm77' is 7C7
2) 6 humans, of whom A and B are two, arrange themselves at random in a row. What is the probability that there are at least 3 humans between A and B?
my go
For question 2, my explanation is:
For 3 between:
[A _ _ _ B] _
Deal with elements between A and B. A and B are gone, 4 are left, so choose 3 out of 4 (4C3) then times by 3! for their arrangements
Arrangements of individual elements (the grouped and single one) = 2!
A and B can switch places = 2
Therefore = 2 x 2! x 4C3 x 3!
For 4 between:
[A _ _ _ _ B]
Elements between A and B = 4C4 x 4!
A and B can switch = 2
Therefore = 2 x 4! x 4C4
what am i doing wrong??
I'd appreciate help with this
(I am not good at this)
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