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A four-digit natural number "abcd" is "balanced" if a + d = b + c. What is the probability that it is a "balanced" if a four-digit natural number is randomly selected?

edited by aboslutelydestroying  Apr 7, 2024
edited by aboslutelydestroying  Apr 7, 2024

#1
+9664
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I could not figure out a nice way to calculate the number of "balanced" numbers, but I wrote a Python program to count them.

The output of the program tells me that there are 615 "balanced" numbers. Hence, the required probability is \(\dfrac{615}{9000} = \dfrac{41}{600}\).

Would love to see a combinatorial approach to this problem.

Apr 8, 2024

#1
+9664
+1

I could not figure out a nice way to calculate the number of "balanced" numbers, but I wrote a Python program to count them.

The output of the program tells me that there are 615 "balanced" numbers. Hence, the required probability is \(\dfrac{615}{9000} = \dfrac{41}{600}\).

Would love to see a combinatorial approach to this problem.

MaxWong Apr 8, 2024
#2
+214
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Thanks