money9111 wrote:how do you know that each can be in each place twice? did you write them out? how do you know to add 3*2, 4*2, and 5*2? it's not so much the math that gets me as much as it's knowing how to set it up.. and go about solving.
I will clarify to you the problem approach.I guess you didn't understand it clearly.
What is the sum of all possible 3-digit numbers that can be constructed using the digits 3,4,5, if each digit can be used only once in each number?
The no. of possible such 3-digit numbers are 3! i.e. 6 numbers.
Now,each no. can only occur for 2 times in units/tens/hundreds place.
so, for units place = 2x(3+4+5) = 24
for tens place= 2x10(3+4+5)=240
for hundreds place = 2x100(3+4+5)=2400
Hence,the sum of the digits will be [spoiler]2400+240+24 = 2664.[/spoiler]
Alternatively,since only 6 numbers are there you can even write it down and solve the question
345
354
435
453
534
543
Quickly,add the no.s and you will get the result.
But if your problem approach is clear,I found the formal method to quicker in this question.
Last edited by
harsh.champ on Sat Feb 06, 2010 1:59 pm, edited 2 times in total.
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