metskj127 wrote:
Bunuel wrote:
What is the sum of all possible 3-digit numbers that can be constructed using the digits 3,4 and 5, if each digit can be used only once in each number?
Three digit number has the form: 100a+10b+c.
# of three digit numbers with digits {3,4,5} is 3!=6.
These 6 numbers will have 6/3=2 times 3 as hundreds digit (a), 2 times 4 as as hundreds digit, 2 times 5 as hundreds digit.
The same with tens and units digits.
100*(2*3+2*4+2*5)+10*(2*3+2*4+2*5)+(2*3+2*4+2*5)=100*24+10*24+24=2664.
Generally the sum of all the numbers which can be formed by using the n distinct digits, is given by the formula:
(n-1)!*(sum of the digits)*(111…..n times)
In our original question: n=3. sum of digits=3+4+5=12. --> (3-1)!*(12)*(111)=24*111=2664.
Hope it's clear.
Hi Bunuel,
Thanks for the help. For the above equation, is 111 constant, or does that number increase to match n? Example, if n=5, would we use 11,111?
(n-1)!*(sum of the digits)*(11,111…..n times)[/b]
Yes, there should be n number of 1's. So, if n=5, it should be 11,111.[/quote]
Is this applicable only when the digits can't be repeated ?
What id we were to form four digit numbers using 3,4,5 with any one being repeated ?
Are there more formulae for such questions ?