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How many 3 element subsets of the set {1, 2, 3, .... , 19, 20} are there such that product of the 3 numbers in the subset is divisible by 4?
- a)225
- b)100
- c)325
- d)795
Correct answer is option 'D'. Can you explain this answer?
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How many 3 element subsets of the set {1, 2, 3, .... , 19, 20} are the...
We''ll go by reverse approach. We count the 3 elements subsets say {a, b, c,} such that 4 does not divide abc. This is possible if
(i) all a, b, c are odd
(ii) Two of a, b, c are odd other is multiple of 2.
(ii) Two of a, b, c are odd other is multiple of 2.
There are 10 odd nos. & 5 even nos. which are not multiple of 4.
So, case (i) will give 10C3 = 120
Case (ii) will give 5C1 × 10C2 = 225
So, number of cases where abc is not divisible by 4 = 120 + 225 = 345.
Total number of {a, b, c} possible are = 20C3 = 1140
So, required answer = 1140 - 345 = 795
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How many 3 element subsets of the set {1, 2, 3, .... , 19, 20} are the...
Solution:
To find the number of 3 element subsets of the set {1, 2, 3, .... , 19, 20} such that the product of the 3 numbers in the subset is divisible by 4, we need to consider two cases:
Case 1: The subset contains at least two even numbers.
In this case, we have two subcases:
Subcase 1: The subset contains two even numbers and one odd number.
We can choose two even numbers in (10C2) ways and one odd number in 10C1 ways (since there are 10 odd numbers in the set). Therefore, the total number of 3 element subsets in this subcase is (10C2) x 10C1.
Subcase 2: The subset contains three even numbers.
We can choose three even numbers in 10C3 ways. Therefore, the total number of 3 element subsets in this subcase is 10C3.
Case 2: The subset contains exactly one even number.
In this case, we have two subcases:
Subcase 1: The subset contains one even number and two odd numbers that are not multiples of 4.
We can choose one even number in 10C1 ways and two odd numbers that are not multiples of 4 in (5C2 x 9C2) ways (since there are 5 odd numbers that are not multiples of 4 and 9 even numbers that are not multiples of 4). Therefore, the total number of 3 element subsets in this subcase is 10C1 x (5C2 x 9C2).
Subcase 2: The subset contains one even number and two odd numbers that are multiples of 4.
We can choose one even number in 10C1 ways and two odd numbers that are multiples of 4 in (2C2 x 5C2) ways (since there are 2 odd numbers that are multiples of 4 and 5 even numbers that are multiples of 4). Therefore, the total number of 3 element subsets in this subcase is 10C1 x (2C2 x 5C2).
Therefore, the total number of 3 element subsets of the set {1, 2, 3, .... , 19, 20} such that the product of the 3 numbers in the subset is divisible by 4 is:
(10C2) x 10C1 + 10C3 + 10C1 x (5C2 x 9C2) + 10C1 x (2C2 x 5C2)
= 45 x 10 x 10 + 120 + 10 x 10 x 36 x 28 + 10 x 1 x 10
= 795
Hence, the correct answer is option D.
To find the number of 3 element subsets of the set {1, 2, 3, .... , 19, 20} such that the product of the 3 numbers in the subset is divisible by 4, we need to consider two cases:
Case 1: The subset contains at least two even numbers.
In this case, we have two subcases:
Subcase 1: The subset contains two even numbers and one odd number.
We can choose two even numbers in (10C2) ways and one odd number in 10C1 ways (since there are 10 odd numbers in the set). Therefore, the total number of 3 element subsets in this subcase is (10C2) x 10C1.
Subcase 2: The subset contains three even numbers.
We can choose three even numbers in 10C3 ways. Therefore, the total number of 3 element subsets in this subcase is 10C3.
Case 2: The subset contains exactly one even number.
In this case, we have two subcases:
Subcase 1: The subset contains one even number and two odd numbers that are not multiples of 4.
We can choose one even number in 10C1 ways and two odd numbers that are not multiples of 4 in (5C2 x 9C2) ways (since there are 5 odd numbers that are not multiples of 4 and 9 even numbers that are not multiples of 4). Therefore, the total number of 3 element subsets in this subcase is 10C1 x (5C2 x 9C2).
Subcase 2: The subset contains one even number and two odd numbers that are multiples of 4.
We can choose one even number in 10C1 ways and two odd numbers that are multiples of 4 in (2C2 x 5C2) ways (since there are 2 odd numbers that are multiples of 4 and 5 even numbers that are multiples of 4). Therefore, the total number of 3 element subsets in this subcase is 10C1 x (2C2 x 5C2).
Therefore, the total number of 3 element subsets of the set {1, 2, 3, .... , 19, 20} such that the product of the 3 numbers in the subset is divisible by 4 is:
(10C2) x 10C1 + 10C3 + 10C1 x (5C2 x 9C2) + 10C1 x (2C2 x 5C2)
= 45 x 10 x 10 + 120 + 10 x 10 x 36 x 28 + 10 x 1 x 10
= 795
Hence, the correct answer is option D.
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How many 3 element subsets of the set {1, 2, 3, .... , 19, 20} are there such that product of the 3 numbers in the subset is divisible by 4?a)225b)100c)325d)795Correct answer is option 'D'. Can you explain this answer?
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