anonymous
  • anonymous
In the integer 2,478 the digits are all different and they increase from left to right. How many integers between 2,000 and 3,000 have digits that are all different and increase from left to right?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
5 210 35 123 14 Answer sheet says it's one of these numbers.... how do i get to them and which one is it
Callisto
  • Callisto
I've a slow method, that is list out all possible numbers... for 2000 -3000 It cannot be a no.. like 20xx, 21xx, nor 22xx So, start with 23xx it can be 2345, 2346, 2347, 2348, 2349 <- x5 2356, 2357,2358, 2359 <-x4 2367, 2368, 2369 <- x3 2378 , 2379 <-x2 2389 <-x1 Total no for 23xx = 5 +4+3+2+1 For 24xx it can be 2456, 2457, 2458, 2459 <- x4 From the above, we can deduce that there are 3+ 2+ 1 more no. can be form for 24xx Total number for 24xx= 4+3+2+1 For 25xx it can be 2567, 2568, 2569 <-x3 From the above, we can deduce that there are 2+ 1 more no. can be form for 25xx Total number for 25xx= 4+3+2+1 So from the above, we can deduce that the total numbers can be formed = (5+4+3+2+1) +(4+3+2+1) +(3+2+1) +(2+1) +1 = 15 +10 + 6 + 3+1 = 35
anonymous
  • anonymous
Thank you, i messed up with the sum but got it! Thank you

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anonymous
  • anonymous
I know the answer is 35. I think my solution is sort of cheating. I hope someone has a better solution. We have 2 < a < b < c. Find all possible permutations of a, b, c whereas a, b, c are integers if a=3, then b can = 4 and c can take on 5 values, 5,6,7,8,9. or b can = 5 can c can take 4 values, 6,7,8,9. Thus, if a=3, bc has 5+4+3+2+1=15 permutations if a=4, bc takes on 4+3+2+1=10 permutations similarly we reach a 15+10+6+3+1=35

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