permutation and combination

Question

A three digit number can be formed using a digits 0,1,2,3,4,5,6,7,8,9 then find no of numbers can formed when repetition of digits not allowed and each 3 digit number divisible by 5

Answer 1

D1  D2 , 5 is divisible by 5. D1 can be any of 1,2,3,4,6,7,8,9 -- 8 possibilities.  D2 can be any of remaining eight digits ( zero can be used now)  thus we have 8.8.= 64 numbers (e.g. 125, 105....).

 

D1  D2 , 0 is alsodivisible by 5. D1 can be any of 1,2,3,4,5,6,7,8,9 -- 9 possibilities.  D2 can be any of remaining eight digits (zero cannot be used now)  thus we have 9.8.= 72 numbers (e.g. 120, 150....).

 

Thus we have a total of 136 such numbers.

 

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Pradip 

Answer 2

,5*(9*8)=360 possibilities divisible by whatever; there are 5 odd digit numbers, so once last digit is chosen (making number odd), you can permutate down from the remaining 9 numbers. Thus there is 8*9 = 72 odd non repeating numbers for each odd digit for three digit numbers. And you can do this five times.

 

For or numbers divisible by 5, you have a possibility of only current restriction ending in number five;

this there here is a total of 72 of these numbers. 

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