Hugoba wrote:
Hi Bunuel,
Thanks for the reply.
It may seem a stupid, but I'm confused by the way the problem was worded
Consider the following set A:
A= {b,b,b,b,b,b,b}
Now consider the following subsets C and D of A:
C={b,b,b,b}; D={b,b,b,b}
C was obtained from A by selecting the underlined elements of A as follows:
A= {b,b,b,b,b,b,b}
D was obtained from A by selecting the underlined elements of A as follows:
D= {b,b,b,b,b,b,b}
Aren't sets C and D (by the purpose of the problem) the same?
I have the same concern as
Hugoba - after research out,
the OA may not be correct if all baguettes are considered the same.
(
Bunuel, kindly correct if my following explanation flawed)
First, back to basic,
the definition of combination:
"A k-combination of a set S is a subset of k distinct elements of S. If the set has n elements, the number of k-combinations is equal to the binomial coefficient or \(C\binom{n}{k}\)"The key thing to note here is the
distinct elements - that means \(C\binom{n}{k}\) formula can only apply to the set if its elements are different, if there are some repetition of elements (or identical elements) in the group,
\(C\binom{n}{k}\) formula cannot be applied.Example: how many different combination of choossing 4 baguettes out of 7 identical baguettes (without considering the order)?
Ans: \(C\binom{7}{4}\)? - Nope!
No matter what way you choose, from left to right, center to left to right, first of last, etc etc - you'll always end with the identical 4 baguettes --> that means there is only
1 expected result from this --> so
the number of different combination is always 1
Back to the problem, how I tackle this?1. You buy 6 out of 10 (7B+3S) (S-for something else) - all baguettes are the same, and all something else are the same.
2. How many possible way you can buy 6 loads?
- only 4 0S + 6B
1S + 5B
2S + 4B --> this is a desirable outcome3S + 3B
So the probability of desirable outcome is
\(p = 1/4 = 0.25\)That's my understanding, hope to hear more from you guys!
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