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

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

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
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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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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.

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.

computer-bot Please explain highlighted portions in red color.

wizardofoddz Please explain portion highlighted in red color.

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

Answer (B)

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