Permutation involving sum of digits

There is a certain variety of questions that involve getting a bunch of numbers using permutation, and then doing some operations on the numbers we get. The questions can get a little overwhelming considering the sheer magnitude of the number of numbers involved! Let’s take a look at that concept today. We will explain it using an example and then take a question as an exercise:   What is the sum of all four digit integers formed using the digits 1, 2, 3 and 4 such that each digit is used exactly once in each integer? First of all, we will use our basic counting principle to find the number of integers that are possible. The first digit can be chosen in 4 ways. The next one in 3 ways since each digit can be used only once. The next one in 2 ways and there will be only one digit left for the last place. This gives us a total of 4*3*2*1 = 24 ways of writing such a four digit number. This is what some of the numbers will look like: 1234 1243 1324 1342 … 2143 … 4321 Now we need to add these 24 integers to get their sum. Note that since each digit has an equal probability of occupying every place, out of the 24 integers, six integers will have 1 in the units place, six will have 2 in the units place, another six will have 3 in the units place and the rest of the six will have 4 in the units place. The same is true for all places – tens, hundreds and thousands. Imagine every number written in expanded form such as: 1234 = 1000 + 200 + 30 + 4 2134 = 2000 + 100 + 30 + 4 …etc. For the 24 numbers, we will get six 1000’s, six 2000’s, six 3000’s and six 4000’s. In addition, we will get six 100’s, six 200’s, six 300’s and six 400’s. For the tens place, will get six 10’s, six 20’s, six 30’s and six 40’s. And finally, in the ones place we will get six 1’s, six 2’s, six 3’s and six 4’s. Therefore, the total sum will be: 6*1000 + 6*2000 + 6*3000 + 6*4000 + 6*100 + 6*200 + … + 6*3 + 6*4 = 6*1000*(1 + 2 + 3 + 4) + 6*100*(1 + 2 + 3 + 4) + 6*10*(1 + 2 + 3 + 4) + 6*1*(1 + 2 + 3 + 4) = 6*1000*10 + 6*100*10 + 6*10*10 + 6*10 = 6*10*(1000 + 100 + 10 + 1) = 1111*6*10 = 66660 Note that finally, there aren’t too many actual calculations, but there is some manipulation involved. Let’s look at a question using this concept now: What is the sum of all four digit integers formed using the digits 1, 2, 3 and 4 (repetition is allowed) A) 444440 B) 610000 C) 666640 D) 711040 E) 880000 Conceptually, this problem isn’t much different from the previous one. Using the same basic counting principle to get the number of integers possible, the first digit can be chosen in 4 ways, the next one in 4 ways, the next one in again 4 ways and finally the last digit in 4 ways. This is what some of the numbers will look like: 1111 1112 1121 … and so on till 4444. As such, we will get a total of 4*4*4*4 = 256 different integers. Now we need to add these 256 integers to get their sum. Since each digit has an equal probability of occupying every place, out of the 256 integers, 64 integers will have 1 in the units place, 64 will have 2 in the units place, another 64 integers will have 3 in the units place and the rest of the 64 integers will have 4 in the units place. The same is true for all places – tens, hundreds and thousands. Therefore, the total sum will be: 64*1000 + 64*2000 + 64*3000 + 64*4000 + 64*100 + 64*200 + … + 64*3 + 64*4 = 1000*(64*1 + 64*2 + 64*3 + 64*4) + 100*(64*1 + 64*2 + 64*3 + 64*4) + 10*(64*1 + 64*2 + 64*3 + 64*4) + 1*(64*1 + 64*2 + 64*3 + 64*4) = (64*1 + 64*2 + 64*3 + 64*4) * (1000 + 100 + 10 + 1) = 64*10*1111 = 711040 So our answer is D.