in Combinatory
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If I have 4 digit number formed using 5,6,7,8. What is the sum of all such 4 digit numbers?

Please provide the detailed solution, I am unable to follow counting problems. Thank You.
in Combinatory
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i am assuming repetition are allowed

see,

number will look like $a\ b\ c\ d$

fixing d as 5, you can get total of $4*4*4=64$ combinations for other 3 positions, now due to symmetry if you fix a,b,c at ones position, you will get also $64$ combinations

same thing will happen for tens, hundredth and thousands position.

now, since there are 4 types of figures at ones position each coming 64 times, sum at ones position=$(5+6+7+8)*64=1664$, 166 will go carry to next place.

for tens position $(5+6+7+8)*64 + 166(carry) = 1830$, 183 will go carry to next place.

for hundreds position $(5+6+7+8)*64 + 183(carry) = 1847$, 184 will go carry to next place

for thousands position $(5+6+7+8)*64 + 184(carry) = 1848$

total sum = 1848704

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2 Answers

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for getting the sum of 4 digits numbers we can apply the formula as follows:-

(sum of given digits) * 1111...n (given number of digits) * (number of digits-1)!

eg:- to find the sum of 4 digits using 5 , 6 , 7 , 8

sol:- (5+6+7+8) * (1111) * ( 4-1 )! =173316

=
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There are $4!=24$ such numbers. We can index them with $i=1,…,24$ and write each of them as

$n_{i}=a_{i}+10b_{i}+100c_{i}+1000d_{i}$

Then

$\sum_{i=1}^{24}n_{i}=\sum_{i=1}^{24}a_{i}+10\sum_{i=1}^{24}b_{i}+100\sum_{i=1}^{24}c_{i}+1000\sum_{i=1}^{24}d_{i}$


$\left\{ i\mid a_{i}=5\right\}, \left\{ i\mid a_{i}=6\right\},\left\{ i\mid a_{i}=7\right\}$ and $\left\{ i\mid a_{i}=8\right\}$

Since it's a $4$ digit number, each digit will appear $6=\frac{24}{4}$ times in each of units, tens, hundreds, and thousands place, So:

$\sum_{i=1}^{24}a_{i}=6.5+6.6+6.7+6.8=6(5+6+7+8)=156$

This can also be applied for $b,c$ and $d$ and finally we find:

$\sum_{i=1}^{24}n_{i}=156+10*156+100*156+1000*156=1111*156=173316$

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