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How many 3 digit integers can be chosen such that none of

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Silver89
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How many 3 digit integers can be chosen such that none of [#permalink] New post   Updated on: 27 Jan 2012, 08:35
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How many 3 digit integers can be chosen such that none of the digits appear more than twice and none of the digits equal zero ?

A. 729
B. 720
C. 648
D. 640
E. 576

what is the wrong of my method which is :
3*(9*9*8)
where first digit can be chosen by 9 ways
second digit can be chosen by 9 ways
third digit can be chosen by 8 ways
and since it's anagram of ( SSD ) it can be arranged in 3 ways so the answer IMO is 3*(9*9*8)= 1944 which is wrong
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B

Originally posted by Silver89 on 27 Jan 2012, 07:42.
Last edited by Silver89 on 27 Jan 2012, 08:35, edited 1 time in total.
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How many 3 digit integers can be chosen such that none of [#permalink] New post   27 Jan 2012, 09:24
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Silver89 wrote:
How many 3 digit integers can be chosen such that none of the digits appear more than twice and none of the digits equal zero ?

A. 729
B. 720
C. 648
D. 640
E. 576


Thanks,
In fact you can solve it in another way too which is
number of ways without restriction = 9*9*9 = 729
number of ways that break restriction = 9*1*9 = 9
so the answer is 729-9 = 720

My question why can't i say
first digit can be chosen by 9 ways
second digit can be chosen by 9 ways
third digit can be chosen by 8 ways
and we can arrange it in 3 ways, thus the answer is 9*9*8*3


Ah, I got your question.

Yes, the easiest way would be as you pointed out:
A. Numbers without restriction = 9^3;
B. 3-digit integers with all alike digits = 9 (111, 222, ..., 999);

A-B=9^3-9=720.

Now, if we go your way it should be 9*8*7+9*8*3=720, where 9*8*7 is # of 3-digit integers with all distinct digits and 9*8*3 is # of 3-digit integers with two alike digits (XXY - 9*8 and multiplied by 3!/2! to get different permutations).

So, as there are two cases you cannot combine them in one formula and apply one factorial correction (3!/2!) to it.

Hope it's clear.
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Re: How many 3 digit integers can be chosen such that none of [#permalink] New post   27 Jan 2012, 08:40
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Thanks,
In fact you can solve it in another way too which is
number of ways without restriction = 9*9*9 = 729
number of ways that break restriction = 9*1*9 = 9
so the answer is 729-9 = 720

My question why can't i say
first digit can be chosen by 9 ways
second digit can be chosen by 9 ways
third digit can be chosen by 8 ways
and we can arrange it in 3 ways, thus the answer is 9*9*8*3
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Re: How many 3 digit integers can be chosen such that none of [#permalink] New post   29 Nov 2014, 01:40
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Bunuel wrote:
Silver89 wrote:
Thanks,
In fact you can solve it in another way too which is
number of ways without restriction = 9*9*9 = 729
number of ways that break restriction = 9*1*9 = 9
so the answer is 729-9 = 720

My question why can't i say
first digit can be chosen by 9 ways
second digit can be chosen by 9 ways
third digit can be chosen by 8 ways
and we can arrange it in 3 ways, thus the answer is 9*9*8*3


Ah, I got your question.

Yes, the easiest way would be as you pointed out:
A. Numbers without restriction = 9^3;
B. 3-digit integers with all alike digits = 9 (111, 222, ..., 999);

A-B=9^3-9=720.

Now, if we go your way it should be 9*8*7+9*8*3=720, where 9*8*7 is # of 3-digit integers with all distinct digits and 9*8*3 is # of 3-digit integers with two alike digits (XXY - 9*8 and multiplied by 3!/2! to get different permutations).

So, as there are two cases you cannot combine them in one formula and apply one factorial correction (3!/2!) to it.

Hope it's clear.



Sorry I don't understand both ways you mentioned. I dont know what is wrong with my way
9 ways for the first digit
9 ways for the second digit
8 ways for the last one
=> 9 x 9 x 8 = 648 ways. please explain! Many thanks
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Re: How many 3 digit integers can be chosen such that none of [#permalink] New post   29 Nov 2014, 05:39
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HuongShashi wrote:
Bunuel wrote:
Silver89 wrote:
Thanks,
In fact you can solve it in another way too which is
number of ways without restriction = 9*9*9 = 729
number of ways that break restriction = 9*1*9 = 9
so the answer is 729-9 = 720

My question why can't i say
first digit can be chosen by 9 ways
second digit can be chosen by 9 ways
third digit can be chosen by 8 ways
and we can arrange it in 3 ways, thus the answer is 9*9*8*3


Ah, I got your question.

Yes, the easiest way would be as you pointed out:
A. Numbers without restriction = 9^3;
B. 3-digit integers with all alike digits = 9 (111, 222, ..., 999);

A-B=9^3-9=720.

Now, if we go your way it should be 9*8*7+9*8*3=720, where 9*8*7 is # of 3-digit integers with all distinct digits and 9*8*3 is # of 3-digit integers with two alike digits (XXY - 9*8 and multiplied by 3!/2! to get different permutations).

So, as there are two cases you cannot combine them in one formula and apply one factorial correction (3!/2!) to it.

Hope it's clear.



Sorry I don't understand both ways you mentioned. I dont know what is wrong with my way
9 ways for the first digit
9 ways for the second digit
8 ways for the last one
=> 9 x 9 x 8 = 648 ways. please explain! Many thanks


We want to find the number of 3-digit integers, without 0, in which none of the digits appear more than twice. So, the number should be XXY (2 digits are alike) or XYZ (all digits are distinct).

Now, {all 3-digit integers, without 0} = {3-digit integers with all the same digits} + {3-digit integers where 2 digits are alike} + {3-digit integers with distinct digits}.

Therefore,
{3-digit integers where 2 digits are alike} + {3-digit integers with distinct digits} = {all 3-digit integers, without 0} - {3-digit integers with the same digits}.

{3-digit integers with 2 digits are alike} + {3-digit integers with distinct digits} = 9^3 - 9.

Hope it helps.
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Re: How many 3 digit integers can be chosen such that none of [#permalink] New post   29 Nov 2014, 08:00
HuongShashi wrote:
Bunuel wrote:
Silver89 wrote:
Thanks,
In fact you can solve it in another way too which is
number of ways without restriction = 9*9*9 = 729
number of ways that break restriction = 9*1*9 = 9
so the answer is 729-9 = 720

My question why can't i say
first digit can be chosen by 9 ways
second digit can be chosen by 9 ways
third digit can be chosen by 8 ways
and we can arrange it in 3 ways, thus the answer is 9*9*8*3


Ah, I got your question.

Yes, the easiest way would be as you pointed out:
A. Numbers without restriction = 9^3;
B. 3-digit integers with all alike digits = 9 (111, 222, ..., 999);

A-B=9^3-9=720.

Now, if we go your way it should be 9*8*7+9*8*3=720, where 9*8*7 is # of 3-digit integers with all distinct digits and 9*8*3 is # of 3-digit integers with two alike digits (XXY - 9*8 and multiplied by 3!/2! to get different permutations).

So, as there are two cases you cannot combine them in one formula and apply one factorial correction (3!/2!) to it.

Hope it's clear.



Sorry I don't understand both ways you mentioned. I dont know what is wrong with my way
9 ways for the first digit
9 ways for the second digit
8 ways for the last one
=> 9 x 9 x 8 = 648 ways. please explain! Many thanks


As bunnel aptly said, it is quite simple to follow two steps :

1) 9*9*9 =729 ;this includes the no , which are repeated more than twice.
2) we have 9 no's excluding 0 ,whose all digits are same.

Hence , total no =729-9=720

Answer B
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Re: How many 3 digit integers can be chosen such that none of [#permalink] New post   28 Mar 2015, 12:00
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Direct with restriction:
Let m = number of three-digit integers with one digit appearing twice
m=9*1*8=72
However, there are C(3,2) ways that this condition can occur.
M=Total Appearing Twice=m*C(3,2)=72*3=216

Let z=number of three-digit integers with each digit appearing once.
z=9*8*7=504
There are C(3,3) ways that this condition can occur
Z=Total Appearing Once=z*C(3,3)=504*1=504

Total satisfying the problem condition=M+Z=216+504=720

B
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Re: How many 3 digit integers can be chosen such that none of [#permalink] New post   29 Mar 2015, 08:32
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Silver89 wrote:
How many 3 digit integers can be chosen such that none of the digits appear more than twice and none of the digits equal zero ?

A. 729
B. 720
C. 648
D. 640
E. 576

what is the wrong of my method which is :
3*(9*9*8)
where first digit can be chosen by 9 ways
second digit can be chosen by 9 ways
third digit can be chosen by 8 ways
and since it's anagram of ( SSD ) it can be arranged in 3 ways so the answer IMO is 3*(9*9*8)= 1944 which is wrong


The simplest way to solve this is to find the numbers without any conditions applied and then from that subtract the numbers the satisfy the conditions.

Total 3-digit numbers without any digit being 0 = 9 * 9 * 9 = 729.
Out of these only 111, 222, 333, ... , 999 violate the given condition that a digit cannot be repeated more than twice.
Hence there are 9 such numbers.
Hence 729 - 9 = 720.

Hence option (B).

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Re: How many 3 digit integers can be chosen such that none of [#permalink] New post   30 Mar 2015, 02:40
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Let's say this number is ABC.

Let's calculate how many numbers with two similar non zero digits and how many numbers with distinct non zero digits:

1. Two similar none zero digits: A has 9 possibilities, B has 1 possibility and C has 8 possibilities. So 9.1*8=72 different numbers. However, this configuration can be modeled in 3 ways, depending on the position of the similar numbers: XXY, XYX, YXX, therefore 72*3=216.

2. Distinct numbers: A has 9 possibilities, B has 8 possibilities and C has 7 possibilities, therfore: 9*8*7=504.

In total we have 504+216=720.
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How many 3 digit integers can be chosen such that none of [#permalink] New post   08 Sep 2016, 20:37
How many 3 digit integers can be chosen such that none of the digits appear more than twice and none of the digits equal zero?

9/9*9/10*9/10=729/900
729-9 triples=720
B. 720
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Re: How many 3 digit integers can be chosen such that none of [#permalink] New post   11 Nov 2017, 03:32
I solved it by 9 * 9 * 8

First digit nine ways
Second digit nine ways since we can repeat the number twice
Third number 8

Why is it wrong?
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Re: How many 3 digit integers can be chosen such that none of [#permalink] New post   11 Nov 2017, 05:10
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santro789 wrote:
I solved it by 9 * 9 * 8

First digit nine ways
Second digit nine ways since we can repeat the number twice
Third number 8

Why is it wrong?


Please read carefully here: https://gmatclub.com/forum/how-many-3-d ... l#p1035258
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Re: How many 3 digit integers can be chosen such that none of [#permalink] New post   04 Jul 2018, 03:11
TooLong150 wrote:
Direct with restriction:
Let m = number of three-digit integers with one digit appearing twice
m=9*1*8=72
However, there are C(3,2) ways that this condition can occur.
M=Total Appearing Twice=m*C(3,2)=72*3=216

Let z=number of three-digit integers with each digit appearing once.
z=9*8*7=504
There are C(3,3) ways that this condition can occur
Z=Total Appearing Once=z*C(3,3)=504*1=504

Total satisfying the problem condition=M+Z=216+504=720

B



Can someone explain the logic behind "There are C(3,3) ways that this condition can occur". I've never seen this.
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How many 3 digit integers can be chosen such that none of [#permalink] New post   23 Sep 2019, 08:00
Silver89 wrote:
How many 3 digit integers can be chosen such that none of the digits appear more than twice and none of the digits equal zero ?

A. 729
B. 720
C. 648
D. 640
E. 576


SOLUTION 1 (CASES)
\(digits={1,2,3,4,5,6,7,8,9}=9\)
\([000:all.different]…9•8•7=72•7=490+14=504\)
\([110:pair.different]…9•1•8•arrangements(3!/2!)=72•3=216\)
\(total=504+216=720\)

Answer (B)

SOLUTION 2 (TOTAL-NOTCASE)
\(digits={1,2,3,4,5,6,7,8,9}=9\)
\([XXX:any.digit]…9•9•9=729\)
\([333:triple]…9•1•1=9\)
\(total-not=729-9=720\)

Answer (B)
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Re: How many 3 digit integers can be chosen such that none of [#permalink] New post   30 Oct 2022, 08:09
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