Question

The sum of all numbers greater than 10,000 by using the digits 0, 2, 4, 6, 8 considering no digit being repeated in any number is
A) 5199970
B) 5199960
C) 5199950
D) 5199940

Answer 1

Firstly, find the sum of all numbers formed using digits 0, 2, 4, 6, 8 and greater than 10000, where 0 can also be at first place.
Then, find the sum of all numbers formed using digits 2, 4, 6, 8.
Finally, subtract the sum of all numbers formed using digits 2, 4, 6, 8 from the sum of all numbers formed using digits 0, 2, 4, 6, 8 and greater than 10000, where 0 can also be at first place, to get the required answer.

Complete step by step solution:
Here, it is given that numbers greater than 10,000 are formed by using the digits 0, 2, 4, 6, 8 considering no digit being repeated.
Now, we need to find the sum of all the above numbers.
To do so, first we need to find the sum of numbers that are greater than 10000.
So, sum of all number formed using digits 0, 2, 4, 6, 8 and greater than 10000, where 0 can also be at first place $= \left( {5 - 1} \right)! \times \left( {0 + 2 + 4 + 6 + 8} \right)\left( {\dfrac{{{{10}^5} - 1}}{{10 - 1}}} \right)$
$= 4! \times 20 \times \left( {\dfrac{{100000 - 1}}{9}} \right) \\\ = 24 \times 20 \times \dfrac{{99999}}{9} \\\ = 480 \times 11111 \\\ = 5333280 \\\$
Now, 0 must not be at first place, because if 0 is at first place the number becomes a 4-digit number.
So, sum of all numbers formed using digits 2, 4, 6, 8 $= \left( {4 - 1} \right)! \times \left( {2 + 4 + 6 + 8} \right)\left( {\dfrac{{{{10}^4} - 1}}{{10 - 1}}} \right)$
$= 3! \times 20 \times \dfrac{{10000 - 1}}{9} \\\ = 6 \times 20 \times \dfrac{{9999}}{9} \\\ = 120 \times 1111 \\\ = 133320 \\\$
Now, to get the sum of all number formed using digits 0, 2, 4, 6, 8 and greater than 10000, we have to subtract sum of all numbers formed using digits 2, 4, 6, 8 i.e. 133320 from sum of all number formed using digits 0, 2, 4, 6, 8 and greater than 10000, where 0 can also be at first place i.e. 5333280.
= 5333280 – 133320
= 5199960

So, option (B) is correct.

Note:
Permutation:
Permutation is the method to arrange all the elements of any set in some sequence or some order. The formula to calculate permutation is ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ .
Combination:
Combination is the method to select some elements from a collection in which order or sequence is not important. The formula to calculate the combination is ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ .
The relation between permutation and combination is ${}^n{C_r} = r! \times {}^n{P_r}$.