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19 Oct 2020, 05:43
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59% (01:34) correct
41%
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wrong
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A university offers 5 different courses. A student must opt for at-least one of the courses. How many different ways the courses can be opted?
a. 5!
b. 5! - 1
c. \(2^5 \)
d. \(2^5\) – 1
e. \(2^6\)
a. 5!
b. 5! - 1
c. \(2^5 \)
d. \(2^5\) – 1
e. \(2^6\)
19 Oct 2020, 06:52
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The total number of ways of choosing everything (including 0) out of n distinct things = \(2^n\)
When choosing at least 1, we remove the option of choosing nothing, which is 1 way. Then the total ways = \(2^n - 1 \)
Here n = 5, therefore the number of ways = \(2^5 - 1\)
Option D
Arun Kumar
When choosing at least 1, we remove the option of choosing nothing, which is 1 way. Then the total ways = \(2^n - 1 \)
Here n = 5, therefore the number of ways = \(2^5 - 1\)
Option D
Arun Kumar
19 Oct 2020, 07:25
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There is an alternate solution.
5c1+5c2+5c3+5c4+5c5=31. To shorten the calculation we can use 5c3=5c2, also 5c4=5c1.
Thanks
Posted from my mobile device
5c1+5c2+5c3+5c4+5c5=31. To shorten the calculation we can use 5c3=5c2, also 5c4=5c1.
Thanks
Posted from my mobile device
