This discussion board is a place to discuss our Daily Challenges and the math and science
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explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.
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Math
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Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3
2×3
2^{34}
234
a_{i-1}
ai−1
\frac{2}{3}
32
\sqrt{2}
2
\sum_{i=1}^3
∑i=13
\sin \theta
sinθ
\boxed{123}
123
Comments
Is the answer B? First you arrange A,E,O, (in that particular order) into 8 space, so 8C3. Then you arrange the final distinct 5 letters without restrictions in the last 5 spaces available, 5!. So 8C3*5!
actually,on my part,....i don't know what's the exact answer.....BUT YOUR DESCRIPTION REGARDING THE QUESTION IS SOUNDING TO BE RIGHT ALONG WITH THE ANSWER TOO....i too thought in the same way and got the answer (B) but also thought that there may be some faults in my thinking....here you are also supporting my thinking procedure....However keep in touch....to see whether others of the friends meet at the same point of argument.........:)
o.k....then your thinking and answer is perfect and transparent.....here you see, others are having same viewpoint and opinion.......thanks Kee W.L........
B.solution is:if we arrange a,e,o into 8 spaces in a particular order , we get 8c3 possible ways.then arrange the remaining numbers with no restriction,so 5! possible ways.so the answer is 8c3*5!
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Is the answer B? First you arrange A,E,O, (in that particular order) into 8 space, so 8C3. Then you arrange the final distinct 5 letters without restrictions in the last 5 spaces available, 5!. So 8C3*5!
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actually,on my part,....i don't know what's the exact answer.....BUT YOUR DESCRIPTION REGARDING THE QUESTION IS SOUNDING TO BE RIGHT ALONG WITH THE ANSWER TOO....i too thought in the same way and got the answer (B) but also thought that there may be some faults in my thinking....here you are also supporting my thinking procedure....However keep in touch....to see whether others of the friends meet at the same point of argument.........:)
thanx for communication....
o.k....then your thinking and answer is perfect and transparent.....here you see, others are having same viewpoint and opinion.......thanks Kee W.L........
I agree with Kee Wei L.
B.solution is:if we arrange a,e,o into 8 spaces in a particular order , we get 8c3 possible ways.then arrange the remaining numbers with no restriction,so 5! possible ways.so the answer is 8c3*5!
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right....:)
ans is b because choose three letters a,e,o from 8 letters as 8c3 and left five letters can be rearranged can be in 5 !
manner
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yes u r right....:)