The digits 1,2,3,4&5 can be arranged to form many different 5 digit positive integers with five distinct digits. How many such integers is the digit 1 to the left of the digit 2? Two such integers to include are 14,352 and 51,234 Please don’t google or use AI
Isnt this an elementary schoo mathl question?
Possibly
I have no clue but if we put enough orders in we can calculate the liquidity in transit before it hits the local market and trade off this to rearrange the amount of numbers in the question from 5 to 69
fix 1 and 2. count the permutations. add them up. 10 ways of fixing 1 and 2. 6 permutations of other numbers for each case. 6 times 10 is 60. now tc or gtfo
12xxx 1x2xx 1xx2x 1xxx2 x12xx x1x2x x1xx2 xx12x xx1x2 xxx12 10 combinations with 6 each = 60
60
1. 21xxx - 6 options: 345, 354, 435, 453, 534, 543 2. 2x1xx - 6 options 3. 2xx1x - 6 options 4. 2xxx1 - 6 options 5. x21xx - 6 options 6. x2x1x - 6 options 7. x2xx1 - 6 8. xx21x - 6 9. xx2x1 - 6 10. xxx21 - 6 60
Switch the 1 & 2 but same result
With 2 at the ith index: 0-0 1-6 2-12 3-18 4-24 Total = 60
5!/2=60 by argument of symmetry. Either 1 is to the left of 2 or 2 is to the left of 1. Both have the same number of cases.
There are 5! possible numbers. Each number is formed by randomly choosing a remaining digit. Half of the time 1 is chosen before 2, so the answer is 5!/2 = 60
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