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by MAAJ » Thu Apr 28, 2011 6:04 am
rakesh8403 wrote:Using all the digits 1,1,1,0,0,3 how many 6-digit numbers can be formed?
I'm no good at this but let me try...

Total digit 6-digit numbers = 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720
6-digit numbers starting with one 0 = 2 x 4 x 4 x 3 x 2 x 1 = 192
6-digit numbers starting with two 0 = 2 x 1 x 4 x 3 x 2 x 1 = 48

720 - 192 - 48 = 480

Is 480 the OA?
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by GMATGuruNY » Thu Apr 28, 2011 6:04 am
rakesh8403 wrote:Using all the digits 1,1,1,0,0,3 how many 6-digit numbers can be formed?
Good arrangements = total possible arrangements - bad arrangements.

Total number of ways to arrange 1,1,1,0,0,3 = 6!/3!2! = 60.

Bad arrangements will put 0 in the 100,000th place.
Number of ways to arrange the remaining 5 digits 1,1,1,0,3 = 5!/3! = 20.

Good arrangements = Total - Bad = 60-20 = 40.
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by vineeshp » Thu Apr 28, 2011 6:11 am
Since it is 6 digit, all numbers have to be used and the two zeros cannot take the first place.

4 * 5 * 4 * 3 * 2 * 1
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Just telling you what I know and think. I am not the expert. :)
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by manpsingh87 » Thu Apr 28, 2011 6:13 am
rakesh8403 wrote:Using all the digits 1,1,1,0,0,3 how many 6-digit numbers can be formed?
left most digit can be either 1 or 3;
case 1) when the left most digit is 1; 1-----; remaining 5 numbers would be 11003; which can arrange themselves in 5!/2!*2!=30
case 2) when the left most digit is 3; 3-----; remaining 5 numbers would be 11100; which can arrange themselves in 5!/3!*2!= 20

hence required no. or numbers= 30+20=50
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by GMATGuruNY » Thu Apr 28, 2011 6:25 am
manpsingh87 wrote:
rakesh8403 wrote:Using all the digits 1,1,1,0,0,3 how many 6-digit numbers can be formed?
left most digit can be either 1 or 3;
case 1) when the left most digit is 1; 1-----; remaining 5 numbers would be 11003; which can arrange themselves in 5!/2!*2!=30
case 2) when the left most digit is 3; 3-----; remaining 5 numbers would be 11100; which can arrange themselves in 5!/3!*2!= 20

hence required no. or numbers= 30+20=50
Perfectly reasoned. Just be careful with the arithmetic:
5!/(3!2!) = 10, yielding 30+10 = 40 possible numbers.
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by rakesh8403 » Thu Apr 28, 2011 6:26 am
though the correct ans is 40 , can somebody explain how he has calculate bad choices ... bacause i was doing

6!/(3!*2!) - 5!/3! -4/3!
= 60 - 20 -4
=36
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by vineeshp » Thu Apr 28, 2011 6:34 am
Dint consider the repeats there. :)
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by manpsingh87 » Thu Apr 28, 2011 6:35 am
GMATGuruNY wrote:
manpsingh87 wrote:
rakesh8403 wrote:Using all the digits 1,1,1,0,0,3 how many 6-digit numbers can be formed?
left most digit can be either 1 or 3;
case 1) when the left most digit is 1; 1-----; remaining 5 numbers would be 11003; which can arrange themselves in 5!/2!*2!=30
case 2) when the left most digit is 3; 3-----; remaining 5 numbers would be 11100; which can arrange themselves in 5!/3!*2!= 20

hence required no. or numbers= 30+20=50
Perfectly reasoned. Just be careful with the arithmetic:
5!/(3!2!) = 10, yielding 30+10 = 40 possible numbers.
thanks a lot sir...!!! really appreciate your help..!!! :D
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