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

Question

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

GMATinsight · Answer

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

Answer: option B

JeffTargetTestPrep · Answer

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.

Answer: B

push12345 · Answer

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

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computer-bot · Answer

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.

wizardofoddz · Answer

Two cases are involved here:
1. 3 digit numbers with each digit appearing only once.
2. 3 digit numbers with 1 digit appearing twice.

1. 3 digit numbers with each digit appearing only once.

__ __ __

First sport can be filled with 9 digits. Second spot can be filled by all but the digit used in the first spot. Third digit can be filled by all but the 2 digits already used in the first and second spots.

9*8*7 = 504 different numbers.

2. 3 digit numbers with 1 digit appearing twice.

Say we fill the first spot with any one of the 9 digits. Now, if this digit were to reappear in the second or third spot, we shall only have 1 way of filling it.That leaves us with the only other remaining spot, which can be filled with rest of the 8 remaining digits.

9*1*8 or 9*8*1 = 72 ways each

Number of arrangements of 2 spots for the repeating digits out of 3 total spots = 3!/2!

Therefore, total number of ways for case 2 is 72*3 = 216

Hence with case 1 and case 2 taken together, we have a total of 504+216 = 720 ways.

Kinshook · Answer

computer-bot  Please explain highlighted portions in red color.

Kinshook · Answer

wizardofoddz  Please explain portion highlighted in red color.

Mohammadmo · Answer

We have 9×9×9=729 numbers which does not consist of zero and 9 numbers which have the same three digits. So,729-9=720
Option B

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exc4libur · Answer

SOLUTION 1
3d-integers without ZERO: 9•9•9=729
3d-integer triplets: 9•1•1=9
3d-integer pairs without zero: 729-9=720

Answer (B)

SOLUTION 2
3d-integers all different without ZERO  : 9•8•7=504
3d-integers all pairs without ZERO  : 9•1•8•arrangements(3!/2!)=72•3=216
3d-integer pairs without zero: 504+216=720

Answer (B)

bumpbot · Answer

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

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

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