How many even 3-digit positive integers can be formed using digits {0,

Question

How many even 3-digit positive integers can be formed using digits {0, 1, 2, 3, 4, 5, 6}, if the repetition of the digits is NOT allowed ?

A. 105
B. 120
C. 162
D. 168
E. 170

GMATinsight · Answer

7 digits to use are {0, 1, 2, 3, 4, 5, 6}

Repetition not allowed

3 digit numbers to be made

Even Numbers to be made

Case 1  : Hundred place filled by any ODD digit from 0 to 6 i.e. 3 ways {1, 3, 5}

Units place can be filled by any one of 4 even digits {0, 2, 4, 6} = 4 ways

Ten's place can be filled by 5 digits now {any digit from 0 to 6 except the ones used at hundreds and Units place} = 5 ways

Total Possible numbers = 3*4*5 = 60

Case 2  : Hundred place filled by EVEN digits {2, 4, 6} from 1 to 6 i.e. 3 ways

Units place can be filled by any one of remaining 3 even digits {0, 2, 4, 6} except the one used at Hundreds place = 3 ways

Ten's place can be filled by 5 digits now {any digit from 0 to 6 except the ones used at hundreds and Units place} = 5 ways

Total Possible numbers = 3*3*5 = 45

Total Numbers = 60+45 = 105

Answer: Option A

Bunuel · Answer

Try similar question here: https://gmatclub.com/forum/how-many-eve ... 25376.html (repetition IS allowed)

IanStewart · Answer

The units digit is even. If it is zero, we have 1 choice for the units digit, and then the hundreds digit can be any of the 6 remaining numbers, and the tens digit any of the 5 remaining numbers. So we have (6)(5)(1) = 30 numbers ending in zero.

If the units digit is one of the other three even numbers, we have 3 choices for the units digit. Then the hundreds digit can be any of the remaining numbers  besides zero , so we have 5 choices, and finally the tens digit can be any of the 5 remaining numbers. So in this case we have (3)(5)(5) = 75 even numbers not ending in zero.

So in total there are 30 + 75 = 105 possible numbers. We need to divide the problem into cases, because the number of choices we have for the hundreds digit is different depending on whether we use zero for the final digit (since the hundreds digit must be nonzero).

Archit3110 · Answer

we can solve ; total - odd ; even
total possible integers ; 6*6*5 ; 180
case 1; unit and hundereds place is odd ; 3*5*2 ; 30
case 2; unit place is odd and hundred is even ; 3*5*3 ; 45
total odd places ; 75
even would be 180-75 ; 105 
OPTION A

agarwalankur · Answer

Case 1 : Assume Unit digit as 0. Hence no. will be even as last digit is 0. Tenth and Hundredth digits can taken any of the 6 nos. without repetition. No. of ways = 5*6*1=30 Ways

Case 2: Assume Tenth digit as 0.Units digit can be selected in 3 ways(2,4,6). Hundredth's digit can be any of the other 5 no.s. No. of ways = 5*1*3=15 Ways

Case 3: In case 1 and 2 we have already taken cases with 0 as a no. Now in this case excluding 0, Unit's digit can be filled in 3 ways -2,4,6. Tenth digit can be filled in 5 ways(All no.s except 0 and Unit's digit) and similarly hundredth's digit can be filled in 4 ways. No. of ways= 4*5*3= 60 ways

Ans: 30+15+60 = 105 Ways OA

Foxtrot109649 · Answer

Dear All I am very new to this forum and trying to figure out my way through some quant questions I came across this one particular question while going through the explanation I was unable to visualize the highlighted portion. is it possible if any one of you can help me break this down further ,

I was okay with this question when repetition rule was allowed

Please find the question below::

How many even 3-digit positive integers can be formed using digits {0, 1, 2, 3, 4, 5, 6}, if the repetition of the digits is NOT allowed??

Answer:

Case 1: Hundred place filled by any ODD digit from 0 to 6 i.e. 3 ways {1, 3, 5}

Units place can be filled by any one of 4 even digits {0, 2, 4, 6} = 4 ways

Ten's place can be filled by 5 digits now {any digit from 0 to 6 except the ones used at hundreds and Units place} = 5 ways??????

Total Possible numbers = 3*4*5 = 60

Case 2: Hundred place filled by EVEN digits {2, 4, 6} from 1 to 6 i.e. 3 ways

Units place can be filled by any one of remaining 3 even digits {0, 2, 4, 6} except the one used at Hundreds place = 3 ways

Ten's place can be filled by 5 digits now {any digit from 0 to 6 except the ones used at hundreds and Units place} = 5 ways

Total Possible numbers = 3*3*5 = 45

Total Numbers = 60+45 = 105

bumpbot · Answer

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How many even 3-digit positive integers can be formed using digits {0,

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How many even 3-digit positive integers can be formed using digits {0,

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