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Manager  Joined: 25 Nov 2009
Posts: 50
Location: India
How many even 4-digit numbers can be formed, so that the  [#permalink]

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Question Stats: 71% (01:46) correct 29% (01:38) wrong based on 79 sessions

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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
Manager  Joined: 24 Mar 2010
Posts: 79

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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?
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Math Expert V
Joined: 02 Sep 2009
Posts: 59182
How many even 4-digit numbers can be formed, so that the  [#permalink]

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jusjmkol740 wrote:
Q1. 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

Q2. 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

Note: Found these questions. But not the OAs. Can anyone help? Thanks in advance.

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: http://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: $$336+784=1120$$.

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 $$aabc$$;

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

TOTAL: $$4320*0.9=3888$$.

Hope it helps.
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Manager  Joined: 25 Nov 2009
Posts: 50
Location: India

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1
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.
Manager  B
Status: Student
Joined: 26 Aug 2013
Posts: 167
Location: France
Concentration: Finance, General Management
Schools: EMLYON FT'16
GMAT 1: 650 Q47 V32 GPA: 3.44
Re: How many even 4-digit numbers can be formed, so that the  [#permalink]

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1
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
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Manager  Joined: 03 May 2013
Posts: 66
Re: How many even 4-digit numbers can be formed, so that the  [#permalink]

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2
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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Re: How many even 4-digit numbers can be formed, so that the  [#permalink]

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