How many even 4-digit numbers can be formed, so that the

Note: the above questions are beyond the GMAT scope.

How many even 4-digit numbers can be formed, so that the numbers are divisible by 4 and no two digits are repeated?
A. 336
B. 784
C. 1120
D. 1804
E. 1936

I believe with "no two digits are repeated" the question means that all 4 digits are distinct.

Number is divisible by 4 if the last two digits form a number divisible by 4.

Therefore last two digits can be:
00;
04;
08;
12
16;
...
96.

Basically multiples of 4 in the range 0-96, inclusive. Multiples of 4 in a range 0-96, inclusive are \(\frac{last \ multiple \ in \ the \ range \ - \ first \ multiple \ in \ the \ range}{4}+1=25\) (for more on this issue: https://gmatclub.com/forum/totally-basic ... ml#p730075).

But 3 numbers out of these 25 are not formed with distinct digits: 00, 44, and 88. Hence the numbers we are looking for can have only 22 endings.

If there is 0 in the ending (04, 08, 20, 40, 60, 80 - total 6 such numbers), then the first and second digit can take 8 and 7 choices each = 56 choices total. As there are 6 numbers with 0 in ending, hence total 6*56=336.

If there is no 0 in the ending (total 22 - 6 with zero = 16 such numbers), then the first digit can take 7 choices (10 - 2 digits in the ending and zero, so total 3 digits = 7, as 4-digit number cannot start with zero) and the second digit can take 7 choices too (10 digits - 3 digits we've already used) = 7*7=49 choices total. As there are 16 numbers without zero in ending, hence total 16*49=784.

TOTAL: \(336+784=1120\).

Answer: C.


How many 4-digit numbers can be formed, so that each contains exactly 3 distinct digits?
A. 1944
B. 3240
C. 3850
D. 3888
E. 4216

As there should be 3 distinct digits in 4, the number will have 2 same digits and other 2 distinct - \(aabc\) (1123, 3477, ...)

\(C^3_{10}=120\) - # of ways to choose 3 distinct digits out of 10;
\(C^1_3=3\) - # of ways to choose which digit will be represented twice;
\(\frac{4!}{2!}=12\) - # of permutation of digits \(aabc\);

\(C^3_{10}*C^1_3*\frac{4!}{2!}=4320\).

Now, out of these 4320 numbers some will start with zero, but 4-digit number cannot start with zero as in this case it becomes 3 digit number. So how many out of these 4320 start with zero? As no digit has any preferences so equal numbers will start with each digit: 1/10 will start with 1, 1/10 with 2, 1/10 with 3 and so on. Thus 1/10=0.1 of 4320 will start with 0 and 0.9 of 4320 will start with digit other than zero.

TOTAL: \(4320*0.9=3888\).

Answer: D.

Hope it helps.