Permutations and Combinations

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The discussion focuses on calculating the sum of all five-digit numbers formed by the digits 1, 2, 3, 4, and 5 without repetition. There are 120 unique permutations of these digits. Each digit appears equally in each place value, with 24 occurrences per digit in each position. The sum of the ones digits totals 360, while the contribution from the tens digits is 3,600. This method allows for an efficient calculation of the overall sum of these five-digit numbers.
Artermis
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If anyone is able to help me with this question regarding introductory Data Management, I would be grateful.

Find the sum of all the five digit numbers that can be formed using the digits 1,2,3,4, and 5 without repeating any digit.

Thank you!

Artermis
 
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Well, if you had one digit,it would be 1. If you had two, it would be two. If you had three digits, 1,2,3; well then we have 123,132,231,213, 312,321 = 6, and so on...
 
Any permutation of the 5 digits will give a different number, so there are 5!=120 numbers in total. If you list them vertically (and mentally ofcourse) you can see that adding them is relatively simple by adding the ones digits, tens digits, etc seperately. There are 4!=24 numbers ending in 1, 24 ending in 2 etc. So the sum of the ones digits is 24(1)+24(2)+..+24(5)=24(1+2+3+4+5)=360
Likewise, the contribution of the tens digits is: 10 x 24(1+2+3+4+5)=3600
etc.
 

Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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