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# How many 3-digit integers can be chosen such that none of the digits a

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Joined: 02 Sep 2009
Posts: 51215
How many 3-digit integers can be chosen such that none of the digits a  [#permalink]

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15 May 2018, 01:32
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Difficulty:

75% (hard)

Question Stats:

50% (02:08) correct 50% (01:49) wrong based on 66 sessions

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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 0?

(A) 729
(B) 720
(C) 648
(D) 640
(E) 576

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

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15 May 2018, 02:33
1
Bunuel 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 0?

(A) 729
(B) 720
(C) 648
(D) 640
(E) 576

Total 3 digit numbers without digit zero = 9*9*9 = 729

Numbers with one of the digits appearing more than twice i.e. thrice = 9 {Numbers like 111, 222, 333, 444, 555, 666, 777, 888, 999}

SO favourable cases = 729 - 9 = 720

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

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16 May 2018, 09:54
Bunuel 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 0?

(A) 729
(B) 720
(C) 648
(D) 640
(E) 576

Let’s first disregard the condition of none of the digits appearing more than twice and count the three-digit numbers where none of the digits is 0. We see that there are 9 choices for each of the digits; therefore, there are 9^3 = 729 such numbers.

Now, we can deal with the condition that none of the digits should appear more than twice. If a digit of a 3-digit number appears more than twice, then it must appear all 3 times and there are only 9 numbers that have this property: 111, 222, …, 999. Thus, out of the 729 numbers, 9 do not satisfy this property and 729 - 9 = 720 do satisfy.

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

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29 May 2018, 06:03
Another approach

When none repeat

9*8*7=504

When two digit repeat
9C1*8C1*3!/2!=216

2! for compensating of two digits same

Total=504+216=720

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

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04 Jul 2018, 02:00
push12345 wrote:
Another approach

When none repeat

9*8*7=504

When two digit repeat
9C1*8C1*3!/2!=216

2! for compensating of two digits same

Total=504+216=720

Give kudos if it helps

Posted from my mobile device

What I don't understand is that why in the first case we've applied the FCP without combinations (i.e., 9*8*7) and for the "two digit repeat" case we've applied the FCP with combinations. I am totally stumped by the calculation for the second case. Kindly elaborate if possible.
Re: How many 3-digit integers can be chosen such that none of the digits a &nbs [#permalink] 04 Jul 2018, 02:00
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