using only the digit 2,3, and 9 how many six digit numbers can be formed which are divisible by 6
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Being a multiple of 6, the number has to be multiple of 3 as well as even number.
The only even digit allowed being 2, it has to be used in the last place. So, the last digit is fixed at 2
Also, being a multiple of 3, the sum of digits must be div by 3: as the digits 3 and 9 are themselves div by 3, hence we should use either three 2s or six 2s (at least one 2 is compulsory, from the previous constraint)
Using six 2s, there is only one number 222222
Using three 2s, the number will be of the form: _ _ _ _ _ 2, where each blank space will be occupied by a digit. Firstly, there will be two more 2s, occupying two of the five blanks, select those in 5C2 = 10 ways. Once done, the remaining three positions (whichever those are) will have two options (3 or 9) each. So, another 2^3 = 8 options. So total of 10*8 = 80 numbers in this scenario
So, we will have 81 possibilities.
The only even digit allowed being 2, it has to be used in the last place. So, the last digit is fixed at 2
Also, being a multiple of 3, the sum of digits must be div by 3: as the digits 3 and 9 are themselves div by 3, hence we should use either three 2s or six 2s (at least one 2 is compulsory, from the previous constraint)
Using six 2s, there is only one number 222222
Using three 2s, the number will be of the form: _ _ _ _ _ 2, where each blank space will be occupied by a digit. Firstly, there will be two more 2s, occupying two of the five blanks, select those in 5C2 = 10 ways. Once done, the remaining three positions (whichever those are) will have two options (3 or 9) each. So, another 2^3 = 8 options. So total of 10*8 = 80 numbers in this scenario
So, we will have 81 possibilities.
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