Question

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?

Answer 1

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

Answer 2

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

Answer 3

Thank you, i messed up with the sum but got it! Thank you

Answer 4

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