How many six digit positive integers that are divisible by 3 can be formed using the digits 0,1,2,3,4,5 and 6 without any of the digits getting repeating?

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Sujoy Das

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Last two digits should be 30,03,12,21,36,63,45,54 to divisible by three... so total possible arrangements could be = 4x(4x3x2x1)x2 = 192 digits
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Answer is 1320 Explaination: To form 6 digit number which is divisilble by 3. first condition is sum of number is divisible by 3i.e., 0+1+2+3+4+5+6=21 which is divisible by 3. since out of 7 digit , only 6 digit has to be placed, one digit has to be removed. (which should be divisible by 3) So, by...
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Answer is 1320 Explaination: To form 6 digit number which is divisilble by 3. first condition is sum of number is divisible by 3i.e., 0+1+2+3+4+5+6=21 which is divisible by 3. since out of 7 digit , only 6 digit has to be placed, one digit has to be removed. (which should be divisible by 3) So, by removing 0 and 3 from series the total number will always be divisible by 3. 1.)) by removing 0 from series(1,2,3,4,5,6):-we get 6 digits can be arranged in 6! times = 720 2.)) by removing 3 from series(0,1,2,4,5,6):- 0 cannot take first place from left so without it, we have remaining 5 number and now including 0 in other 5 places we can arrange 5 digits in 5! ways. = 5*5!= 5*120=600 3)) In total 720+600=1320 read less
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here i have posted the answer.
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1. If '0' is not used then it can be arranged in 6 possible ways 6!=720 numbers 2.if we don't use '3' and '0' cannot be the first digit, hence we have only 5 digits as 1st number they are 1,2,4,5,6 Then remaining 5 digits can be arranged in other 5 places hence we will have 5 digits in 5 ways 5*5!=600...
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1. If '0' is not used then it can be arranged in 6 possible ways 6!=720 numbers 2.if we don't use '3' and '0' cannot be the first digit, hence we have only 5 digits as 1st number they are 1,2,4,5,6 Then remaining 5 digits can be arranged in other 5 places hence we will have 5 digits in 5 ways 5*5!=600 numbers By adding 1 and 2 we will have 720+600=1320 numbers read less
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ans is 1920 divisibility test of 3 is the sum of digits should be divisible by 3. So sum of 0,1,2,3,4,5,6 is 21. case 1: when 1,2,3,4,5,6 numbers are taken possible number will be factorial 6=120 Case 2: when 0,1,2,4,5,6 are taken possible numbers will be 5*5! (as 0 cannot come as first digit otherwise...
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ans is 1920 divisibility test of 3 is the sum of digits should be divisible by 3. So sum of 0,1,2,3,4,5,6 is 21. case 1: when 1,2,3,4,5,6 numbers are taken possible number will be factorial 6=120 Case 2: when 0,1,2,4,5,6 are taken possible numbers will be 5*5! (as 0 cannot come as first digit otherwise that would not be an 6 digit no. Case3 : when 0,1,2,3,4,5 numbers are taken possible number will be 5*5! Ans is 120+(2*5*5!) read less
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ones place can be filled in 6 different ways tens place can be filled in 5 different ways hundredth place can also be filled in 5 different ways ans should be 6!*5!*5!
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What is the value of 6P3?
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