Too Much
How many positive integers have no repeated digits?
The answer is 8877690.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
3 solutions
I think answer will be infinity unless you say about two digit number like that
Log in to reply
No man,think about this,how can a 11 digit number have all distinct digits?
Log in to reply
Oh yeah thanks buddy!
Log in to reply
@Department 8 – No problem man!
@Department 8 – Look up the pigeonhole principle ;)
Yeah Exactly Same Way.
Log in to reply
How did you compute that?That is where I got stuck.
Log in to reply
Calculator
what does "P" stand for?
Log in to reply
Permutation, as in 7P5 is "Line up 5 items out of 7" and is calculated by 7!/(7-5)!
sHOULDNT THE QUESTION Specify the no. of digits the number can atmost have? It would have been better that way, I feel.
Log in to reply
By the Pigeonhole principle, it is impossible for a 10-or-more digit number to have no repeated digits. If you use every digit once, you can only make a 9-digit number. Knowing this was part of the problem.
This is a trick question. Too many positive integers have no repeating digits.
I'm having a little trouble interpreting the original math formula syntax.
I get the same result by looping the 'Non-Repeating permutations formula' from 1 to 10, and summing all results. P(n,k) = n! / (n-k)!
n=10 k = 10..1
Excel syntax Num non repeat digits count =FACT(10)/FACT(10-10) 10 3,628,800 =FACT(10)/FACT(10-9) 9 3,628,800 =FACT(10)/FACT(10-8) 8 1,814,400 =FACT(10)/FACT(10-7) 7 604,800 =FACT(10)/FACT(10-6) 6 151,200 =FACT(10)/FACT(10-5) 5 30,240 =FACT(10)/FACT(10-4) 4 5,040 =FACT(10)/FACT(10-3) 3 720 =FACT(10)/FACT(10-2) 2 90 =FACT(10)/FACT(10-1) 1 10 total 9,864,100
result less 10% of leading zeros
8,877,690
I guess my question is does this approach illustrate the long-hand version of all steps in the math formula of the op's answer?
post-edit: Looks like I lost my table formatting here
I think there is a bit of problem in the solution.There are 10 digits(0 to 9) to constitute a multi-digit number.Thus, there will be 10! solutions for a 10-digit number. Then, for a 9-digit number, 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 solutions.For, an 8-digit number, 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 solutions. Thus, following in this pattern, we get all the number of numbers for all digits. We add them up.However, 0 and 00 and 000 are the same.So, for each digit number, we have taken 0 common.So, subtract 10 from the result. The answer should be 9864090.
Log in to reply
Keep in mind that the first digit of a number cannot be 0, so the first digit has 9 options. There are just 9*9! solutions for a 10-digit number.
Log in to reply
Oh, yeah, you are right!Sorry for the mistake!Upvoted your solution. Very good problem, anyway!
Log in to reply
@Anibrata Bhattacharya – Is 0 a positive integer?
Log in to reply
@Kushagra Sahni – Ohhh!!! Yeah, good take on that! I'll edit the comment.Seriously, why don't I first read the question properly?!
@Kushagra Sahni – It is even, but it is neither positive nor negative.
We can only have 10 digits without repeating ourselves: 0 1 2 3 4 5 6 7 8 9. But wait, only positive numbers, right? Okay, so 1 2 3 4 5 6 7 8 9.
If we have one digit we have no repeating numbers, so we'd have 9 digits here.
Two digits: In the first place we have a choice of 9 digits, and for the second we have 8 choices (because we don't want to repeat ourselves, right?)
Three digits: You see where this is going... 9 x 8 x 7.
Do this upto ten digits. Sum. You tha man.
for the second we have 9 choices ( now 0 is back in the mix)
Your approach to problem is absolutely right ( but you have taken those permutatioms wrongly as you havnt considerex zero for places other than 1st place)
Only up to ten digits can you have distinct digits. For the first digit 0 is not an option, so there are 9 possible digits. For the next digit 0 is an option, therefore there are 9 options. After that the options decrease by one. We start adding the ten digit numbers, then the nine digit numbers, then the eight digit numbers and so on. So the sum is:
( 9 × 9 × 8 × 7 . . . × 2 × 1 ) + ( 9 × 9 × 8 × 7 . . . × 2 ) + ( 9 × 9 × 8 × 7 . . . × 3 ) + . . . = n = 0 ∑ 9 9 × n ! 9 ! = 8 , 8 7 7 , 6 9 0
Starting with as low as a single digit to as most as a 10 digit number, the sum is i = 0 ∑ 9 ( 9 × 9 P i )