What is the sum of all such possible numbers given the following conditions?

Question

What is the sum of all such possible numbers given the following conditions that A 4 digit number is formed using the digits 0,2,4,6,8 without repeating any one of them.?

MyApproach:

If you fix 8 as the last digit, you see that there are 3⋅3⋅2 ways to complete the number. Thus, 8 appears 18 times as the last digit.

By the same logic, if we enumerate all possible numbers using these 5 digits, each number appears 18 times in each of the 3 positions.

At unit Place if we place 8 we can form 4⋅3⋅2=24 times as the unit place

In total, we have (2+4+6+8)(18)(111)+(2+4+6+8)(24)(1000) +(0)(24)(111)=519960

Is my Approach correct towards the problem?Can Anyone guide me if there exist a shorter approach?

Are you sure you are not counting numbers starting with zero? — uniquesolution, Oct 10 '15 at 11:33
@uniquesolution I can fill it in 3.3.2.1(thousand digit can be filled in 3 ways 0 cannot be there to make a 4 digit number )this condition will be fulfilled when we have unit digit as 2,4,6,8.For 0 unit digit we have 24 ways to form the last digit number. — justin takro, Oct 10 '15 at 11:39
@uniquesolution Thankx for guiding.I think i get the right Ans. — justin takro, Oct 10 '15 at 11:55

true blue anil · Answer 1 · 2015-10-10T16:49:01.030

1

An easy way to compute is [sum of $4$ digit strings] - [sum of $3$ digit numbers]

Any column total will be a multiple of $(0+2+4+6+8) = 20$,

For a 4 digit string chosen from $5$ digits, the multiple will be $\dbinom54\times4!/5 = 24$,

and for a $3$ digit number chosen from $4$ digits, the multiple will be $\binom43\times3!/4 = 6$,

hence desired sum = $24\cdot20\cdot1111 - 6\cdot20\cdot111 = 519,960$

edited Oct 10 '15 at 16:49

answered Oct 10 '15 at 11:57

true blue anil

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What is the sum of all such possible numbers given the following conditions?

1 Answers1