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udaypratapsingh99
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Updated on: 12 Jan 2019, 10:52
Originally posted by udaypratapsingh99 on 12 Jan 2019, 09:15.
Last edited by udaypratapsingh99 on 12 Jan 2019, 10:52, edited 2 times in total.
Last edited by udaypratapsingh99 on 12 Jan 2019, 10:52, edited 2 times in total.
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A regiment consisting of 5184 soldiers was asked to march in a rehearsal parade with various possible combinations of rows and columns. What is the probability that in a certain combination, number of rows will be less than the number of columns?
a. \(\frac{1}{2}\)
b. \(\frac{17}{35}\)
c. \(\frac{17}{36}\)
d. \(\frac{17}{18}\)
e. \(\frac{15}{34}\)
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a. \(\frac{1}{2}\)
b. \(\frac{17}{35}\)
c. \(\frac{17}{36}\)
d. \(\frac{17}{18}\)
e. \(\frac{15}{34}\)
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udaypratapsingh99
Joined: 12 Jan 2019
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12 Jan 2019, 09:21
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factorize 5184; \(2^6 * 3^4\)
thus no. of factors = (6+1)*(4+1) =35
And 17 such pairs of rows and column.
so probability = \(\frac{17}{35}\)
thus no. of factors = (6+1)*(4+1) =35
And 17 such pairs of rows and column.
so probability = \(\frac{17}{35}\)
12 Jan 2019, 09:53
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UdayPratapSingh99
How did you get this? Thanks in advance!
