i guess its easy , but me not able to crack

[rakesh8403]

Using all the digits 1,1,1,0,0,3 how many 6-digit numbers can be formed?

[MAAJ]

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?

[GMATGuruNY]

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.

[vineeshp]

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

[manpsingh87]

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

[GMATGuruNY]

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.

[rakesh8403]

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

[vineeshp]

Dint consider the repeats there.

[manpsingh87]

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..!!!