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- Thread starter jacks
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- Feb 5, 2012

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The Total Sum of $5$ Digit no. which can be formed with the Digit $0,1,2,3,4,5,6,7$.

[a] when repetition of digit is allowed

when repetition of digit is not allowed

Hi Jacks,

I'll answer the first part of your question. By the same logic you should be able to complete the second part.

Note that we cannot have zero as the last digit (counting digits from right to left) of the number. Hence for the last digit we have only 7 possibilities. Suppose if we fix 0 as the first digit then we have \(7 \times 8^3=3584\) numbers, and the sum of the first digits of all those numbers will add up to \(0\times 7\times 8^3=0\). Similarly if we take 1 as the fist digit the sum of all those numbers will add up to \(1\times 7\times 8^3=3584\). Generally, if we add up all the ones places of the numbers the sum would be,

\[3584\times (0+1+2+3+4+5+6+7)\]

Similarly if we add up all the tenth places of the numbers the sum would be,

\[3584\times (0+10+20+30+40+50+60+70)=35840\times (0+1+2+3+4+5+6+7)\]

Continuing with this reasoning we get the total sum as,

\[(3584+35840+358400+358400+8^4)\times (0+1+2+3+4+5+6+7)=21288960\]

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(It also include a $4$ Digit no. bcz $0$ at extreme left. )

So Total no.,s e $ \displaystyle = \binom{8}{5} \times 5! = \frac{8!}{3!}$Each digit has an equal chance of being selected into any particular possition.

So, for example, the digit 2 will occupy the units position in $\displaystyle \frac{1}{8}$ of the arrangements.

So The sum of the arrangement each of $5$ position is $\displaystyle = \frac{1}{8}\left(0+1+2+3+4+5+6+7\right)\cdot \frac{8!}{3!}$

So Total Sum of $5$ Digit no. including $0$ at Thousant place is $\displaystyle = \frac{1}{8}\left(0+1+2+3+4+5+6+7\right)\cdot \frac{8!}{3!}\cdot (1111)$

Now We Will Calculate Sum of $4$ Digit no. Using $1,2,3,4,5,6,7$

Using same Idea We Get $\displaystyle = \frac{1}{7}.\frac{7!}{3!}\cdot \left(1+2+3+4+5+6+7\right)\cdot (1111)$

So Total Sum of $5$ Digit no. is $\displaystyle = \frac{1}{8}\left(0+1+2+3+4+5+6+7\right)\cdot \frac{8!}{3!}\cdot (1111) - \frac{1}{7}.\frac{7!}{3!}\cdot \left(1+2+3+4+5+6+7\right)\cdot (1111)$

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