
The number of 4 digit numbers divisible by 3 that can be formed by 4 different even digits {0, 2, 4, 5, 6, 8} is?
A. 36
B. 60
C. 30
D. 24
Answer
572.7k+ views
Hint: For a number to be divisible by 3, the sum of the digits of the number must be a multiple of 3. So find the sets of four digits from {0, 2, 4, 6, 8} (only even digits) which when added gives a multiple of 3. And for a 4-digit number, the leftmost digit must not be a zero (0).
Complete step by step solution:
We are given to find the number of 4 digit numbers divisible by 3 that can be formed by 4 different even digits {0, 2, 4, 5, 6, 8}. Given that the 4-digit numbers should have even digits.
5 is an odd number so we must not include. So the digits are 0, 2, 4, 6, 8. From these 5 digits, there are 16 4-digit numbers possible totally without having the leftmost digit as 0.
But only two sets of 4 digits gives a multiple of 3 when the digits are added, and those are {0, 2, 4, 6} and {0, 4, 6, 8}
So in the first set, {0, 2, 4, 6}, the leftmost digit should not be zero so it can be selected from the remaining 3 numbers {2, 4, 6} and the remaining 3 digits can be selected from the remaining 3 numbers.
In the same way, in {0, 4, 6, 8}, the leftmost digit should not be zero so it can be selected from the remaining 3 numbers {4, 6, 8} and the remaining 3 digits can be selected from the remaining 3 numbers.
Total no. of 4 digit numbers which are divisible by 3 are $ {}_{}^3C_1^{} \times {}_{}^3C_1^{} \times {}_{}^2C_1^{} \times {}_{}^1C_1^{} + {}_{}^3C_1^{} \times {}_{}^3C_1^{} \times {}_{}^2C_1^{} \times {}_{}^1C_1^{} $
$ \to 3 \times 3 \times 2 + 3 \times 3 \times 2 = 18 + 18 = 36 $
Therefore, 36 4-digit numbers are possible.
So, the correct answer is “Option A”.
Note: While selecting an object from a group of objects combinations must be used. When the order of the objects matter, permutations must be used or else combinations. Here we need not follow a strict order to arrange the digits so we have used combinations. The formula of $ {}_{}^nC_r^{} $ is $ \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $
Complete step by step solution:
We are given to find the number of 4 digit numbers divisible by 3 that can be formed by 4 different even digits {0, 2, 4, 5, 6, 8}. Given that the 4-digit numbers should have even digits.
5 is an odd number so we must not include. So the digits are 0, 2, 4, 6, 8. From these 5 digits, there are 16 4-digit numbers possible totally without having the leftmost digit as 0.
But only two sets of 4 digits gives a multiple of 3 when the digits are added, and those are {0, 2, 4, 6} and {0, 4, 6, 8}
So in the first set, {0, 2, 4, 6}, the leftmost digit should not be zero so it can be selected from the remaining 3 numbers {2, 4, 6} and the remaining 3 digits can be selected from the remaining 3 numbers.
In the same way, in {0, 4, 6, 8}, the leftmost digit should not be zero so it can be selected from the remaining 3 numbers {4, 6, 8} and the remaining 3 digits can be selected from the remaining 3 numbers.
Total no. of 4 digit numbers which are divisible by 3 are $ {}_{}^3C_1^{} \times {}_{}^3C_1^{} \times {}_{}^2C_1^{} \times {}_{}^1C_1^{} + {}_{}^3C_1^{} \times {}_{}^3C_1^{} \times {}_{}^2C_1^{} \times {}_{}^1C_1^{} $
$ \to 3 \times 3 \times 2 + 3 \times 3 \times 2 = 18 + 18 = 36 $
Therefore, 36 4-digit numbers are possible.
So, the correct answer is “Option A”.
Note: While selecting an object from a group of objects combinations must be used. When the order of the objects matter, permutations must be used or else combinations. Here we need not follow a strict order to arrange the digits so we have used combinations. The formula of $ {}_{}^nC_r^{} $ is $ \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $
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