Sum of the digits.
The digits in the number 1143 are arranged in all possible ways to form distinct 4 digit integers. What is the sum of all these 4 digit distinct integers? Can some one come up with a simple solution for the above problem?
Thanks.
First let me simplify the
First let me simplify the question a bit How many distinct 4 digit numbers can be made of the numbers 1143 with out repetition. There are many ways of approaching such questions but the simplest is to use the following formula: "The number of permutations of n elements with n1 of repeated element, n2 of another repeated element, etc. taken n at a time. " P=n!/(n1!)(n2!) ( please check http://cat4mba.com/math-e-book/permutations for the details) So here n=4 i.e four different numbers and 1 is repeated twice so n1=2 Thus P=4!/2! = 12 So 12 different numbers can be made. ( Just to verify I am writing these numbers (please don’t do so in any exam) 1143, 1134, 3114, 4113, 3411, 4311, 1413, 1314, 1341, 1431, 3141, 4131) Now let’s find out what is the sum of all the possible numbers We have 12 different numbers i.e 12 numbers with a uint place, 10th place, 100th place and 1000th place. In all these places the digits will be distributed proportionally i.e In unit place we will have 6 1’s , 3 3’s and 3 4’s So the sum is 6*1 + 3*3 + 3 * 4 = 6 + 9 + 12 =27 Thus Sum of all the places 27 + 27*10 + 27*100 + 27 * 1000 = 27 * 1111 = 29997.