Hello,
For the following:
How many odd positive integers n are there such that 40,000 < n < 100,000 and n is made up of the digits 0, 1, 2, 3, 4 and 5?
A) 1296
B) 1295
C) 750
D) 749
E) 150
OA: A
I was trying to solve as follows:
The number of digits are 5.
The last digit could be 1 or 3 or 5. Hence, we have 3 choices for the last digit.
The first digit has to be 4 or 5. Hence, we have 2 choices for the first digit.
Now, the digit could be say 41,111 or 41,113 or 41,114 or 41,235
Hence, the number of integers are: ( 2 x 6 x 6 x 6 x 3 )/ ( 3!)
I am dividing by 3! to avoid double counting the repeating digit in the 2nd, 3rd and 4th place.
However, I think my answer is wrong since the OA is different.
Can you please assist?
Thanks,
Sri
Finding the number of odd positive integers
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Number of options for the ten-thousands digit = 2. (4 or 5.)gmattesttaker2 wrote: How many odd positive integers n are there such that 40,000 < n < 100,000 and n is made up of the digits 0, 1, 2, 3, 4 and 5?
A) 1296
B) 1295
C) 750
D) 749
E) 150
OA: A
Number of options for the thousands digit = 6. (0, 1, 2, 3, 4, or 5.)
Number of options for the hundreds digit = 6. (0, 1, 2, 3, 4, or 5.)
Number of options for the tens digit = 6. (0, 1, 2, 3, 4, or 5.)
Number of options for the units digit = 3. (1, 3, or 5.)
To combine these options, we multiply:
2*6*6*6*3 = 1296.
The correct answer is A.
Your solution is perfect until the portion in red.I was trying to solve as follows:
The number of digits are 5.
The last digit could be 1 or 3 or 5. Hence, we have 3 choices for the last digit.
The first digit has to be 4 or 5. Hence, we have 2 choices for the first digit.
Now, the digit could be say 41,111 or 41,113 or 41,114 or 41,235
Hence, the number of integers are: ( 2 x 6 x 6 x 6 x 3 )/ ( 3!)
I am dividing by 3! to avoid double counting the repeating digit in the 2nd, 3rd and 4th place.
We would divide by 3! if the order of the 3 middle digits didn't matter -- if 41235 were considered the SAME result at 42315.
But 41235 is a DIFFERENT integer from 42315.
Thus, there is no reason to divide by 3!.
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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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Hello Mitch,GMATGuruNY wrote:Number of options for the ten-thousands digit = 2. (4 or 5.)gmattesttaker2 wrote: How many odd positive integers n are there such that 40,000 < n < 100,000 and n is made up of the digits 0, 1, 2, 3, 4 and 5?
A) 1296
B) 1295
C) 750
D) 749
E) 150
OA: A
Number of options for the thousands digit = 6. (0, 1, 2, 3, 4, or 5.)
Number of options for the hundreds digit = 6. (0, 1, 2, 3, 4, or 5.)
Number of options for the tens digit = 6. (0, 1, 2, 3, 4, or 5.)
Number of options for the units digit = 3. (1, 3, or 5.)
To combine these options, we multiply:
2*6*6*6*3 = 1296.
The correct answer is A.
Your solution is perfect until the portion in red.I was trying to solve as follows:
The number of digits are 5.
The last digit could be 1 or 3 or 5. Hence, we have 3 choices for the last digit.
The first digit has to be 4 or 5. Hence, we have 2 choices for the first digit.
Now, the digit could be say 41,111 or 41,113 or 41,114 or 41,235
Hence, the number of integers are: ( 2 x 6 x 6 x 6 x 3 )/ ( 3!)
I am dividing by 3! to avoid double counting the repeating digit in the 2nd, 3rd and 4th place.
We would divide by 3! if the order of the 3 middle digits didn't matter -- if 41235 were considered the SAME result at 42315.
But 41235 is a DIFFERENT integer from 42315.
Thus, there is no reason to divide by 3!.
Thanks a lot for your detailed explanation. It is clear now.
Best Regards,
Sri