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How many 3 element subsets of set X = {1, 2, 3....20} can be formed such that the product of the 3 numbers in the subset is divisible by 4?
Correct answer is '795'. Can you explain this answer?
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How many 3 element subsets of set X = {1, 2, 3....20} can be formed s...
Total number of subsets with 3 elements = 20C3 = 1140
Total number of subsets = total number of subsets where the product of 3 numbers in the subset divisble by 4 + total number of subsets where the products of 3 numbers in the subset is not divisble by 4
We shall find the total number of subsets where the products of 3 numbers in the subset is not divisble by 4
For the given set, product of 3 three numbers is not divisble in 2 cases:
Case 1: when all the three numbers are odd
Case 2: when two numbers are odd and third number is a even number which is not a multple of 4
Case 1: In the given set, there are 10 odd numbers
The number of ways where all the three numbers are odd = 10C3 = 120
Case 2: There are 5 even numbers in the given set which are not multiples of 4.
The number of ways where two numbers are odd and a even number which is not multiple of 4 = 10C2 X 5C1 = 225
Total = 120 + 225 = 345
Hence, the total number of subsets where the product of 3 numbers in the subset divisble by 4 = 1140 – 345 = 795
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How many 3 element subsets of set X = {1, 2, 3....20} can be formed s...
To find the number of 3-element subsets of set X = {1, 2, 3....20} such that the product of the 3 numbers in the subset is divisible by 4, we need to consider the different cases and possibilities.
Case 1: Number of even elements in the subset is 3
In this case, we have 10 even numbers in set X (2, 4, 6, 8, 10, 12, 14, 16, 18, 20). We need to choose 3 numbers from these 10 even numbers.
Number of ways to choose 3 numbers from 10 = C(10, 3) = (10!)/(3!*(10-3)!) = 10*9*8/(3*2*1) = 120
Case 2: Number of even elements in the subset is 2
In this case, we have 10 even numbers in set X and we need to choose 2 even numbers and 1 odd number from these 10 even numbers and 10 odd numbers respectively.
Number of ways to choose 2 numbers from 10 even numbers = C(10, 2) = (10!)/(2!*(10-2)!) = 10*9/(2*1) = 45
Number of ways to choose 1 number from 10 odd numbers = C(10, 1) = 10
Total number of ways in this case = 45 * 10 = 450
Case 3: Number of even elements in the subset is 1
In this case, we have 10 even numbers in set X and we need to choose 1 even number and 2 odd numbers from these 10 even numbers and 10 odd numbers respectively.
Number of ways to choose 1 number from 10 even numbers = C(10, 1) = 10
Number of ways to choose 2 numbers from 10 odd numbers = C(10, 2) = (10!)/(2!*(10-2)!) = 10*9/(2*1) = 45
Total number of ways in this case = 10 * 45 = 450
Case 4: Number of even elements in the subset is 0
In this case, we have 10 odd numbers in set X and we need to choose 3 odd numbers from these 10 odd numbers.
Number of ways to choose 3 numbers from 10 odd numbers = C(10, 3) = (10!)/(3!*(10-3)!) = 10*9*8/(3*2*1) = 120
Total number of ways in this case = 120
Total number of ways
Adding up the number of ways from all the cases, we get:
120 + 450 + 450 + 120 = 1140
But some of these subsets have odd products, which are not divisible by 4. We need to remove those subsets from the total.
Subsets with odd products
To find the number of subsets with odd products, we need to consider the following cases:
Case 1: All 3 numbers are odd
In this case, we have
Case 1: Number of even elements in the subset is 3
In this case, we have 10 even numbers in set X (2, 4, 6, 8, 10, 12, 14, 16, 18, 20). We need to choose 3 numbers from these 10 even numbers.
Number of ways to choose 3 numbers from 10 = C(10, 3) = (10!)/(3!*(10-3)!) = 10*9*8/(3*2*1) = 120
Case 2: Number of even elements in the subset is 2
In this case, we have 10 even numbers in set X and we need to choose 2 even numbers and 1 odd number from these 10 even numbers and 10 odd numbers respectively.
Number of ways to choose 2 numbers from 10 even numbers = C(10, 2) = (10!)/(2!*(10-2)!) = 10*9/(2*1) = 45
Number of ways to choose 1 number from 10 odd numbers = C(10, 1) = 10
Total number of ways in this case = 45 * 10 = 450
Case 3: Number of even elements in the subset is 1
In this case, we have 10 even numbers in set X and we need to choose 1 even number and 2 odd numbers from these 10 even numbers and 10 odd numbers respectively.
Number of ways to choose 1 number from 10 even numbers = C(10, 1) = 10
Number of ways to choose 2 numbers from 10 odd numbers = C(10, 2) = (10!)/(2!*(10-2)!) = 10*9/(2*1) = 45
Total number of ways in this case = 10 * 45 = 450
Case 4: Number of even elements in the subset is 0
In this case, we have 10 odd numbers in set X and we need to choose 3 odd numbers from these 10 odd numbers.
Number of ways to choose 3 numbers from 10 odd numbers = C(10, 3) = (10!)/(3!*(10-3)!) = 10*9*8/(3*2*1) = 120
Total number of ways in this case = 120
Total number of ways
Adding up the number of ways from all the cases, we get:
120 + 450 + 450 + 120 = 1140
But some of these subsets have odd products, which are not divisible by 4. We need to remove those subsets from the total.
Subsets with odd products
To find the number of subsets with odd products, we need to consider the following cases:
Case 1: All 3 numbers are odd
In this case, we have
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How many 3 element subsets of set X = {1, 2, 3....20} can be formed such that the product of the 3 numbers in the subset is divisible by 4?Correct answer is '795'. Can you explain this answer?
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