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

Question

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

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

iambroke · Answer

2) its 3888

The number of combinations is 648 but you can arrange them in 6 ways, so 648x6 = 3888 ways..

Thanks

BTW, where did you get these questions from?

Bunuel · Answer

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.

jusjmkol740 · Answer

Bunnel, thnx. After going thru' ur explanation, now I know why u said in the first place that these are beyond GMAT scope. 
Thanks anyway.

Paris75 · Answer

This is how I proceed for those two :

First one;

I used a simple methode:

- You choose the number of integers divided by 4 between 0 and 1000. This number is 250 (use the formula) 
- You multiply this number by 9 (because your number cannot start with a 0, so it is between 1000 and 9999). 
- You find 2250. But you are seeking for distinct numbers, so you need to divided you results by 2 (in order to be divded by 4 a number needs to have it last TWO digits divided by four such as 36 12...), which gives 1125.

You can now chose answer C! (approximation method)

for 2:

- For the first number you can chose 1 among 9 (without the zero) than you can chose 9 again (adding back the zero) than 8.
- You need then to chose one 3 digit among the 3. 
- You have therefore 9*9*8*3. But you need to multiply by 4 (permutation). But watch out, you can end with a zero as a primary digit! So what do you do? You divide by 2 (to avoid those two cases when the first digit = 0)
- Final number 3888.

Hope it helps

vipulgoel · Answer

Alt way

1. 4 digit different, multiple of 4, 4 digit numbers ?? 
calculate the even , four digit differ number, - - - - , 8 * 8 * 7 * 5, { (unit, 0,2,4,6,8) , thousand ( not 0 and one already come at unit)/2 = 1120
why /2 because every second no in even series is multiople of 4

2. - - - - 9 * 9 * 8 * 1 * 6 { start from left, at thousand, 9 digits (not included 0) , at hundred 9 (0 included) , at tens 8, at unit 1( same digit)
why multiplied with 6 - same digits (1,2;13;14;23;24;34) OR we may apply ( 4!/2)/2 why again divided by 2 again because if for example in 1st and 2 2nd and 2nd and 1st are same will creat same no

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

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