How many different number less than 600,000 - Permutation

Question

I need to solve this task - Any help / direction would help me a lot.

"How many different number less than 600,000 can be formed with digits 1,2,3,4,5,6; no digits repeated."

I tried to solve it by using this formula:

Using all digits (no repeats): P(6) - P(5) = 6! - 5! =  500
Removing last digit 6 because 6xx xxx can't be used: 500 - P(5) = 500 - 5! = 380

I need this to be solved using formula (P/C). Any youtube video explaining this case or similar would be great!

With your first formula, why do you have the "-P(5)" term? Also, 6!-5! is not 500. — GB supports the mod strike, Jan 10 '18 at 02:58
My bad, 6!-5! is 600 indeed. Well I substract -P(5) because to get no repeated digits, you do P(n) - P(n-1) no? — Ilija Lončarević, Jan 10 '18 at 03:18
No, using P(n) already takes care of "no repeated digits". 6! = 6 choices for the first digit, x 5 choices for the second digit (the five digits you haven't already used), x 4 choices for the third, and so on. — GB supports the mod strike, Jan 10 '18 at 03:46
By the way, do you have to use all the digits for each number, or can you make numbers like 123? I think we've all been assuming the former, but I just realized this wasn't actually stated. — GB supports the mod strike, Jan 10 '18 at 03:51

score 2 · Answer 1 · answered Jan 10 '18 at 04:27

2

The question doesn't ask how many six digit numbers, it only specifies numbers

The answer should thus be $_6P_1 +\; _6P_2 +\; _6P_3 +\; _6P_4 +\; _6P_5 + \frac56(_6P_6) =1836$

The $\frac56$ in the last term is because you can't use $6$ as the fist digit for a six digit number $< 600000$

answered Jan 10 '18 at 04:27

true blue anil

41,295

So others are wrong? This is an exam question so I was wondering the solution was so easy..? – Ilija Lončarević Jan 11 '18 at 00:57
Oh, I replied to your comment by mistake under Geoffrey Brent's answer. All to the good, he has added a solution based on the interpretation here. – true blue anil Jan 11 '18 at 07:29

score 1 · Answer 2 · answered Jan 10 '18 at 03:16

1

Number of ways to fill the first position is five (using either 1,2,3,4,5), then we will have remaining five positions to fill with five digits ie, $5!$ ways. Therefore total ways are multiplication of $5*5!= 600$.

answered Jan 10 '18 at 03:16

markovchain

441

GB supports the mod strike · Accepted Answer · 2018-01-11T05:05:09.293

0

One way to approach it:

Assuming all digits must be used:

There are 5 possible positions where the 6 can be placed. Once we've placed that, there are 5 positions for the 5 (6 places, minus one that's already been filled), then 4 positions for the 4, and so on. So we end up with 5*5*4*3*2*1 = 5*5! or equivalently 6!-5!.

This is almost identical to the answer given by markovchain, but instead of working from "what digit shall we assign to each position?" it works from the other side, assigning a position to each digit.

In this particular case, either approach is equally easy to use. But for some problems, it may be easier to go from one direction than the other, so worth keeping in mind that both options exist.

Assuming that we don't have to use all digits:

Same as above, but we also need to calculate the totals for 5, 4, 3, 2, and 1 digits. In these cases the "less than 600000" requirement isn't an issue, so any digit can go in any place.

This then gives a total of (6!-5!)+(6*5*4*3*2)+(6*5*4*3)+(6*5*4)+(6*5)+6 solutions.

edited Jan 11 '18 at 05:05

answered Jan 10 '18 at 03:56

GB supports the mod strike

1,359

Thanks! Briefly explained, helped me understand better! – Ilija Lončarević Jan 10 '18 at 03:57
(I feel like markovchain deserves answer credit for this one, mine was posted as a variation on theirs) – GB supports the mod strike Jan 10 '18 at 04:00
I would certainly interpret the wording "How many different numbers less than 600,000....." to include all permissible numbers less than 600,000. The answers you have received from others is assuming that only 6 digit numbers are to be counted, but this is nowhere stated in the question. Geoffrey Brent has also commented on this. – true blue anil Jan 11 '18 at 04:12
@trueblueanil Yes, I've added that interpretation to my answer; the OP's formula got me assuming that all digits must be used, and it took a while to realise that this isn't entirely clear. (I think your interpretation is most likely right, in which case my original answer was incorrect, but there's some room to interpret "with digits" as "using all of the digits".) – GB supports the mod strike Jan 11 '18 at 05:08

How many different number less than 600,000 - Permutation

3 Answers3