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No of possible ways = 4*4! = 96 - shouldn't this be subtracted by the number of combination that has 0 at the start because the the number is technically 4 digits long!

No of possible ways = 4*4! = 96 - shouldn't this be subtracted by the number of combination that has 0 at the start because the the number is technically 4 digits long!

Please, can you explain rijul007

The first spot has 4 possibilities (1,2,4,5). Then, after that spot has been chosen, 4 possibilities remain for the next spot, and so forth.

Re: How many 5 digit numbers can be formed which are divisible [#permalink]

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06 Sep 2013, 06:58

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To form a 5 digit number from available 6 digits 0,1,2,3,4,5 Case 1: Exclude 0 and form a 5 digit number from 1,2,3,4,5..since the sum of these digits is multiple of 3 as per divisibility rule for 3..A number is divisible by 3 if sum of the digits is divisible by 3 hence all 5 digit numbers formed from 1,2,3,4,5 are divisible by 3 5 digit number formed using 1,2,3,4,5 and divisible by 3 is 5! ways=120 Case 2:Include 0 and form a 5 digit number from 0,1,2,3,4,5 possible ways (0,1,2,3,4) (0,1,2,3,5)(0,1,2,4,5)(0,1,3,4,5)(0,2,3,4,5)...of which only a 5 digit number formed from (0,1,2,4,5) is divisible by 3 Since we need a 5 digit number...first digit can be selected from (1,2,4,5) but not 0..which is 4 ways rest of the digits can be selected from the remaining 4 digits..in 4! ways So no of possible ways will be 4*4! ways

Total number of ways to form a 5 digit no divisible by 3 is 5!+4*4!=120+96=216 ways

How many 5 digit numbers can be formed which are divisible by 3 using the numerals 0, 1, 2, 3, 4, 5 (WITHOUT REPETITION)

A. 216 B. 3152 C. 240 D. 600 E. 305

First step:

We should determine which 5 digits from given 6, would form the 5 digit number divisible by 3.

We have six digits: 0, 1, 2, 3, 4, 5. Their sum=15.

For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 (divisible by 3) only 5 digits good to form our 5 digit number would be 15-0={1, 2, 3, 4, 5} and 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3.

Second step:

We have two set of numbers: 1, 2, 3, 4, 5 and 0, 1, 2, 4, 5. How many 5 digit numbers can be formed using these two sets:

1, 2, 3, 4, 5 --> 5! as any combination of these digits would give us 5 digit number divisible by 3. 5!=120.

0, 1, 2, 4, 5 --> here we can not use 0 as the first digit, otherwise number won't be any more 5 digit and become 4 digit. So, desired # would be total combinations 5!, minus combinations with 0 as the first digit (combination of 4) 4! --> 5!-4!=4!(5-1)=4!*4=96

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