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How many three-element subsets of {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} are t
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13 Aug 2023, 23:11
13 Aug 2023, 23:11
Expert Reply
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Question Stats:
83% (02:03) correct
16% (00:58) wrong based on 6 sessions
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How many three-element subsets of {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} are there such that not all three numbers are prime?
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How many three-element subsets of {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} are t
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08 Oct 2023, 09:01
08 Oct 2023, 09:01
Expert Reply
OE
We need to make three element subsets from 10 numbers.
In sets, order of elements doesn’t matter.
As we need to figure out three-element subsets such that not all three numbers are
prime
That means some can be prime.
So, lets figure out the total number of three element subsets and remove the number of
subsets with all three prime numbers.
Hence, we apply Combination.
Total number of 3 element subsets = \(C^{10}_3\)
But the question doesn’t want the sets where all 3 numbers are prime. Prime digits in
the given set are: 2, 3, 5 and 7.
Let’s calculate the number of 3 element sets where all the digits are prime = \(𝐶^3_4\)
Therefore, \(𝐶^10_3- 𝐶^4_3\), will give the number of 3 element subsets from the given 10 numbers that won’t include all prime element set.
\(C^{10}_3= 10!/3!7! = 120\)
(Select 3 from 4 prime)
\(C^4_3=\frac{4!}{3!1!}=4
\)
120-4=116
We need to make three element subsets from 10 numbers.
In sets, order of elements doesn’t matter.
As we need to figure out three-element subsets such that not all three numbers are
prime
That means some can be prime.
So, lets figure out the total number of three element subsets and remove the number of
subsets with all three prime numbers.
Hence, we apply Combination.
Total number of 3 element subsets = \(C^{10}_3\)
But the question doesn’t want the sets where all 3 numbers are prime. Prime digits in
the given set are: 2, 3, 5 and 7.
Let’s calculate the number of 3 element sets where all the digits are prime = \(𝐶^3_4\)
Therefore, \(𝐶^10_3- 𝐶^4_3\), will give the number of 3 element subsets from the given 10 numbers that won’t include all prime element set.
\(C^{10}_3= 10!/3!7! = 120\)
(Select 3 from 4 prime)
\(C^4_3=\frac{4!}{3!1!}=4
\)
120-4=116
How many three-element subsets of {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} are t
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17 Jan 2024, 06:44
17 Jan 2024, 06:44
1
This basically talks about selection of groups of \(3\) digits out of \(10\) digits.
Now order is not important when it comes to forming these subsets of \(3\) digits out of a set of \(10\) digits.
So it is a combination problem.
The number of subsets that can be formed are \(3C10\)
\(3C10 = \dfrac{10!}{3!7!} = 120\)
But we are asked about the number of three-element subsets whose members are not all prime.
The easiest way of doing this is to find out the number of three-element subsets all of whose elements are prime, and subtract it from the total number of three-element subsets that can be formed.
Now there are four prime numbers - \(2,3,5,7\) and we can form \(3C4\) three-element subsets whose elements are all prime.
\(3C4 = \dfrac{4!}{3!1!} = 4\).
Thus the number of three-element subsets whose members are not all prime is \(120-4 = 116\)
Now order is not important when it comes to forming these subsets of \(3\) digits out of a set of \(10\) digits.
So it is a combination problem.
The number of subsets that can be formed are \(3C10\)
\(3C10 = \dfrac{10!}{3!7!} = 120\)
But we are asked about the number of three-element subsets whose members are not all prime.
The easiest way of doing this is to find out the number of three-element subsets all of whose elements are prime, and subtract it from the total number of three-element subsets that can be formed.
Now there are four prime numbers - \(2,3,5,7\) and we can form \(3C4\) three-element subsets whose elements are all prime.
\(3C4 = \dfrac{4!}{3!1!} = 4\).
Thus the number of three-element subsets whose members are not all prime is \(120-4 = 116\)
