How many different positive integers can be formed using each digit in

Question

How many different positive integers can be formed using each digit in the set {0, 1, 3, 3, 3, 7, 8} exactly once?

A. 160
B. 720
C. 1440
D. 4320
E. 5040

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chetan2u · Accepted Answer

There are 7 digits out of which there are three 3s. 
Different ways these can be arranged is 7!/3!, as three 3s can be arranged in 3! Ways we divided total by 3!
7!/3!=7*6*5*4=840.
However 840 is not in choices, and we have to use each digit once. But the numbers starting with 0 actually do not use 0 for example 0133378 is same as 133378, but without digit 0. Thus the question means 7-digit positive integers, and in this case we have to remove numbers starting with 0. 
If first digit is 0, the remaining 6 can be arranged in 6!/3!=6*5*4=120
Total =840-120=720

OTHER WAY 
The first digit can be taken by any of the digits except 0, so 7-1=6 ways. The remaining 6 positions can be arranged in 6!ways. —-6*6!
But there are 3 similar digits so total ways =6*6!/3!=6*6*5*4=720

B

TarPhi · Answer

Bunuel   chetan2u   GMATinsight  can you please tell me why we divide by 3!?
I've always been confused by this :/

GMATinsight · Answer

Since we have three 3's and their arrangement with each other won't make any difference in outcome so we need to exclude the arrangements of those three 3's

Arrangement of those three 3's can be done in 3! ways which is unnecessarily included (in multiplication form) in 6!, so we exclude the effect of those not-required arrangements of three 3's by dividing 6! by 3!

i.e. Arrangements of six digits {1, 3, 3, 3, 7, 8} = 6!/3!

Similarly, Arrangements of letters of word OHIO = 4!/2!

Similarly, Arrangements of letters of word MISSISSIPPI = 11!/(4!*4!*2!)

I hope this helps!
 TarPhi

Kritisood · Answer

hi  chetan2u  in the explanation for the other way, I have a small query. we first take 6 for all nos without 0 and then we multiply this by 6!/3! to account for the remaining 6 digits and the three threes. but what if the first digit we chose would've been a three? then we would only need to divide by 2! and not 3!?

chetan2u · Answer

Hi we find total ways and these are 6*6! if all were different. 
But since three 3s are same, we divide total ways by 3!...6*6!/3!

Don’t confuse it as 6*(6!/3!), it is (6*6!)/3!

Harsh9676 · Answer

Arrangements problem.
IMO 720

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EP2620 · Answer

chetan2u : I'm not sure if I understand this problem correctly!

If an integer can be used only once, my understanding was that we essentially need to find out number of positive integers that can be formed using {0,1,3,7,8}

Calculating all possible integers and subtracting those starting with zero, we get:

1 digit integers = 4
2 digit integers = 5P2 - 4 = 16 
3 digit integers = 5P3 - 4P2 = 48
4 digit integers = 5P4-4P3 = 96
5 digit integers = 5! - 4! = 96

Total possible integers by using each digit  only once  = 260

Where am I going wrong in my approach?

chetan2u · Answer

Hi, 
It doesn’t tell us that digits are different, but it tells us that we have to use EACH digit in the set, so if 3 is given three times, we will have to use digit 3 three times.

ansarose · Answer

I got the answer but I did it differently..don't know if my interpretation is right .. 
If we group all 3 then there will be five digits in the number .this can be arranged in 5! Ways . Now the 3 threes can be arranged in 3! Ways .
Combining these we get 5!*3! Which is 720

Posted from my mobile device

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How many different positive integers can be formed using each digit in

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