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How many 3 digit integers can be chosen such that none of [#permalink]

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27 Jan 2012, 07:42

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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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Last edited by Silver89 on 27 Jan 2012, 08:35, edited 1 time in total.

Re: How many 3 digit integers can be chosen such that none of [#permalink]

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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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How can i lose my faith in life's fairness when i know that the dreams of those who sleep on the feathers are not more beautiful than the dreams of those who sleep on the ground? - Jubran Khaleel Jubran

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.

Re: How many 3 digit integers can be chosen such that none of [#permalink]

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29 Nov 2014, 01:40

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

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.

WE: General Management (Non-Profit and Government)

Re: How many 3 digit integers can be chosen such that none of [#permalink]

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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.

Re: How many 3 digit integers can be chosen such that none of [#permalink]

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28 Mar 2015, 12:00

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

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.

Re: How many 3 digit integers can be chosen such that none of [#permalink]

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30 Mar 2015, 02:40

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.

Re: How many 3 digit integers can be chosen such that none of [#permalink]

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05 Oct 2017, 12:42

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