What is the number of five digit number that can be formed using the

Hi Brian123 , thanks for asking and I'll try my best to break down what I'm doing.

Mr. Bunuel probably had a typo as it doesn't make sense to write 5 twice, he just meant the digits 1 through 9.

Besides that, after we pick 3 unique numbers, lets say 1,2, 3, you face two more choices/problems. Who gets the single copy and who gets two copies? We can have 11223, 12233, or 11233 after selecting those three digits. Furthermore, the same 5-digit combination can create different numbers, 11223 can be arranged as 31122, 31212, 32121, 13122 etc, you get the idea. So if we account for all of those options you'll see how my solution tackles each problem in order.

For the 5-digit combinations, we can use 9C3 * 3 as I just showed for every 3-digit selection there are 3 possible 5-digit combinations. This turns out to be the same as 36*7.

You are right. Edited the typo. Thank you!

Can someone tell me my mistake here?

I went for ways to choose 3 numbers from 1-9 -> 9C3

Then I tried to fill the 5 digit slots->
1. For the non-repeated number, we have 5 spots
2. For the repeated number, we have 4 and 3 spots (and then divide it by 2! since the order of the same digit wont matter)
3. For the other repeated number, we have 2 and 1 spot (again, divide it by 2! since the order of the same digit wont matter)

so 9C3 * 5 *4*3*2*1 / (2! * 2!) = 2520

Hi wishmasterdj,

This problem bugged me an awful lot as well, and I was initially stuck on the same approach as yours, so let me give it a shot!
Firstly, let's use the letters "AABBC" to represent the two sets of repeated digits (A & B) as well as the single digit (C).

I believe your approach is best summarized as the following:
1. Find the (unordered) number of possible combinations of 3 digits from a set of 9 digits. \((9C3 = \frac{9!}{3!*6!} = 84)\)
2. Find the number of configurations of the letters AABBC. \((\frac{5!}{2!*2! = 30})\)
This is what you did when filling the 5 slots.

So far you've chosen 3 numbers (ex. "123") and you've ordered the letters (ex. "AABBC"), but you also need to consider that the final number will change depending on how you assign the numbers to the letters.
Ex. Assigning ABC = 123 (1 assigned to A, 2 assigned to B, 3 assigned to C) is going to give you the number 11223, while assigning ABC = 321 is going to give you the number 33221.

3. The missing (but also flawed) step: Find the number of ways you can slot the 3 different numbers into the 3 different letters, aka. "how many ways can you order 3 items". \(3! = 6\)

Unfortunately, this approach is flawed, because when you multiply these numbers together you end up with \(84*30*6 = 15120\), twice the actual amount.
This issue arises because A and B share the same properties (both digits are represented twice).
Example of this issue:
Case 1
Chose three numbers: 123
Order the 5 digits : AABBC
Assign the 3 digits : ABC = 123
Result = 11223

Case 2
Chose three numbers: 123
Order the 5 digits : CCBBA
Assign the 3 digits : ABC = 321
Result = 11223

Different configurations, same result.


To tackle this issue, use the method mentioned by TestPrepUnlimited:
1. Select the ordered digits, this step both selects the 3 digits and assigns them to the three letters.
Use 9C2 to treat A and B as identical, you're doing this because A and B have the same properties. \((9C2 = \frac{9!}{2!*7!} = 36)\)
That leaves you with 7 possible digits for C, leading to the equation: \(9C2 * 7 = \frac{9!}{2!*7!} * 7 = 36 * 7\)
2. Find the number of configurations of the letters AABBC. \((\frac{5!}{2!*2!} = 30)\)

Final calculations = \(36*7*30 = 7560\)


I hope this made sense to you, I consider myself quite decent at combinatorics, but this one bugged me for a while.

In this question you have 2 numbers occurring the same number of times i.e twice and 1 number occurring once.

You can choose the one number occurring once in 9C1 = 9 ways and the 2 numbers occurring twice in 8C2 = 28 ways and then rearrange all these numbers in (5! / 2!2!) = 30 ways. Total number of ways= 9 * 28 * 30 = 7560

You can choose the two numbers occurring twice in 9C2 = 36 ways and the one number occurring once in 7C1 = 7 ways and then rearrange all these numbers in (5! / 2!2!) = 30 ways. Total number of ways= 36 * 7 * 30 = 7560

You can also choose on number which will occur twice in 9C1 = 9 ways, the next number occurring twice in 8C1 ways and the number occurring once in 7C1 ways.

What we need to remember here is that 2 of these numbers are occurring twice and hence we have to divide by 2! = 2 ways to avoid double counting.

Total number of ways = (9 * 8 * 7/2) * 30 = 7560 ways



Option E

Arun Kumar

Thanks, this really helps! Appreciate you taking out the time.

I am just curious though, whether this is indeed a 600 level question?!

Hi All

I do understand the logic behind \(9C2 * 7 * \frac{5!}{2!*2!}\)

Generally when there are such question related to digits i work on them by following method. However when I tried the same, it is not working

for example: 44117

hence \( 9*1*8*1*7*\frac{5!}{2!*2!}\)

I am not understanding why this is double the correct answer. What is the difference in the meaning \(9C2\) and \(9C1*8C1\). What am i missing here.

Suppose the question had been : find the number of 3 digit numbers with digits 1 2 3 4 5 6 7 8 9 where 2 digits are same and 1 is different..

then it would have been \(9*1*8 * \frac{3!}{2!}\)

But why same approach is not working here.

TestPrepUnlimited chetan2u Bunuel

Hi rsrighosh !

The method you used here seems correct however let's just take two cases before arranging the digits:

44117 and 11447, you included both of these with 9*1*8*1*7. Yet we need to scramble the digits later, so 44117 and 11447 are effectively the same set of numbers, and we don't want to count the same numbers twice. So we should divide by 2 to get the correct count or follow the methods below.

To avoid this duplication we can use 9C2 to represent the two digits that have a pair (9C2 = the number of ways we can choose 2 digits out of 9) then multiply by 7 for the last digit. Or we can see if we pick any 3 digits, we can have AABBC, ABBCC, or AABCC before scrambling the digits. Thue 9C3 * 3 is another method.

Hope this makes sense!

TestPrepUnlimited



Why 44117 and 11447 are considered same set. Aren't they 2 different numbers? Or i understood the question wrong. I read the question thrice just to make sure if i missed something but i still understand that we need to find number of 5 digit numbers, not unique set. Is 14147 is also in the same set?

rsrighosh

With the method I am using, we start by picking out the selection of digits that satisfy the criteria (9C2 * 7), and then randomize the digits after we pick the selections of 5 digits (5!/(2!*2!)). 11447 and 44117 represent "sets" from my first step and each set contains 30 numbers by randomizing their digits. In the first step of selecting the 5 digits, if we pick 11447 and 44117, the numbers that follow are:

11447: 11447, 11474, 11744, 14471, 14741 ..... 30 numbers in total.

44117: 44117, 44171, 44711 .... again 30 numbers in total.

Obviously, the digits are identical so it's the same 30 numbers in both sets. We want to avoid counting the same number twice so we need to divide by 2 somewhere. I used 9C2 * 7 instead of 9*8*7 so that I'm picking unique sets.

TestPrepUnlimited

Makes sense now. Thanks for detailed explanation..

What you are doing wrong is that 9*1*8*1*7*5!/2!2! is arranging the integers twice.
9*8*7 is used when order is important, and 5!/2!2! again gives you choosing ways to arrange 5 numbers out of which two digits are used twice.
You cannot arrange integers twice

So choose three numbers in 9C3 ways or 84 ways. Each of these 84 ways can have any 2 of the 3 digits repeated twice so 84*3C2=252 ways.
Now each set of numbers so chosen can be arranged in 5!/2!2!=30 ways

Total =252*30=7560

There are various ways to do this question as also shown above.

One more way would be

Step I : Choose the three digits that will be used to make the integer
We can choose 3 digits out of 9 in 9C3 ways =\(\frac{9*8*7}{3*2}=84\)

Step II : After you have chosen the three digits as shown in step I, you will now choose two of these three digits twice and the remaining digit, once
So choose 2 out of 3 in 3C2 or 3 ways.

Step III : After we have chosen the five digit, we have to arrange these
Now, the five digits have a pair of 2 digits, so the ways to arrange them =\(\frac{5!}{2!2!}=30\)

Total = 84*3*30=7560

E

hi Bunuel Can you please validate this solution:

I am using SELECTION*PLACING
- - - - -

1st (number selected out of 9 number) that will be used only once and I got 5 empty position to place it : 9C1 * 5C1

2nd (number selected of 8 numbers) now will be used twice and I got 4 empty position to place first of two :: 8C1 * 4C1
Second of two will be the same number so no choice in selection and I got 3 empty position to place :: 1 * 3C1
( will divide by 2! as repeating twice )

3rd (number selected of 7 numbers) now will be used twice and I got 2 empty position to place first of two :7C1 * 2C1
Second of two will be the same number so no choice in selection and I got 1 empty position to place : 1 * 1
(will divide by 2! as repeating twice )

So final : (9C1 * 5C1 * 8C1 * 4C1 * 1 * 3C1 * 7C1 * 2C1 * 1 * 1)/ 2!2! = 7560

Asked: What is the number of five digit number that can be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 in which one digit appears once and two digits appear twice (e.g 41174 is one such number but 75355 is not.) ?

Number of ways to select 3 distinct digits = 9C3 = 84
Number of ways to select 1 digit not repeated = 3C1 = 3
Number of ways to arrange 5 digits = 5!/2! 2! = 30

The number of five digit number that can be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 in which one digit appears once and two digits appear twice = 84*3*30 = 84*90 = 7560

IMO E

In your alternate solution, could u please explain me the logic behind dividing the product of 9*8*7 by 2.Though i get that there are 2 numbers occurring twice,but i still don't get why it has to be divided by 2.

I believe we are dividing by 2 not because the numbers are occurring twice, but because two such pairs exist. It is easier to see this by restricting ourselves to digits 1,2 and 3 (instead of 1..9):
11_22_3
11_33_2
22_11_3
22_33_1
33_11_2
33_22_1
So, we have 11_22_3, but we will also have 22_11_3. Since we are planning to calculate permutations later, we will need to divide by 2 to avoid overcounting.