Permutation Series
What is the sum of all the 4 digit numbers that can be formed with the digits 2 , 3 , 4 , 5 such that there is no repetitions?
The answer is 93324.
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5 solutions
Same method (+1). It can be generalized by any numbers that have to be formed from any number of positive integers without repitition that the sum of the numbers is
number of digits
1
1
⋯
1
1
×
(
Sum of ths digits
×
Number of digits
Total numbers that can be formed
)
.
But note that this method does not apply when the digits contain a 0, cuz 0 cant come as the first digit of the number.
PS. All digits have to be used exactly once.
Same pinch!
2,3,4,5 all appear in the ones, tens, hundreds, and thousands digit 3! times. Thus, in every digits place: 6(2+3+4+5) = 84.
ones digit = 84, carry 8 and leaving 4
tens digit = 84 + 8 = 92, carry 9 and leaving 2
hundreds = 84 + 9 = 93, carry 9 and leaving 3
thousands = 84 + 9 = 93
Hence, 93324
Perfect solution (+1)
The total numbers formed using these digits are 4! = 24. Take the lowest possible number + the highest possible number; the second lowest number + the second highest lowest number; etc... E.g. 2345 + 5432 = 7777 2354 + 5423 = 7777 Since there are 24 numbers, then you could get 12 sums in total. => 7777 * 12 = 93324
Thats one coincidence. Note that it does not apply everytimecuz the numbers formed does not necessarily have to be in arithmetic progression.
Each digit will occur at each of the four places (thousandths, hundredths, tens and ones) exactly 6 times, and these places have place values of 1 0 0 0 , 1 0 0 , 1 0 and 1 respectively, so the sum is 6 ⋅ ( 5 0 0 0 + 5 0 0 + 5 0 + 5 + 4 0 0 0 + 4 0 0 + 4 0 + 4 + 3 0 0 0 + 3 0 0 + 3 0 + 3 + 2 0 0 0 + 2 0 0 + 2 0 + 2 ) = 9 3 3 2 4
Yeah, thats perfect.
2345+5432=7777 2354+5423=7777 Pair nth smallest number with nth largest number and you'll always get sum of 7777. There are 24 numbers so 12 sums of 7777 which yields answer
Thats one coincidence. Note that it does not apply everytimecuz the numbers formed does not necessarily have to be in arithmetic progression.
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Really can you please elaborate more bc any time I've seen problems like these this solution always works.
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Ok, then try a counter example (one from my old tests). Find the sum of the 4 -digit numbers that can be formed from the digits 1,2,4,6 such that each digit occurs exactly once. By your way, you get ( 1 2 4 6 + 6 4 2 1 ) × 2 4 ! = 7 6 6 7 × 1 2 = 9 2 0 0 4 . But, the answer key gives 8 6 6 5 8 and in fact it is correct because the digits given do not follow an arithmetic progression. You got the same solution because the digits given were in arithmetic progression (2,3,4,5 common difference=2). But the digits in this question are not in AP. For the best approach refer to @Rishabh Cool 's solution above.
Again, if one of the digits was zero, the scenario gets more complex. I guess that solves the doubt.
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@Ashish Menon – Ahhhh okay that makes sense. Thank you Ashish!
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@Ashish Sacheti – Nice to know that it makes sense, you're welcome Ashish!
The total numbers formed using these digits are 4 ! = 2 4 . And note that the technique I would be using to calculate the sum is to calculate sum at one's place, ten's place , hundred's place and so on respectively and then interpret as a 1 a 2 a 3 a 4 = 1 0 0 0 a 1 + 1 0 0 a 2 + 1 0 a 3 + a 4 .
Sum at one's place: Note that it is equally likely for 2 , 3 , 4 , 5 to appear at one's place, hence they would appear 4 4 ! = 6 times. Hence sum = 6 ( 2 + 3 + 4 + 5 ) = 6 ( 1 4 ) = 8 4 .
By symmetry Sum at ten's, hundred and thousand's place will be same i.e 8 4 .
Hence the answer would be :
1 0 0 0 ( 8 4 ) + 1 0 0 ( 8 4 ) + 1 0 ( 8 4 ) + 1 ( 8 4 )
= 8 4 ( 1 0 0 0 + 1 0 0 + 1 0 + 1 ) = 8 4 ( 1 1 1 1 ) = 9 3 3 2 4