How many different five-digit codes can be picked from the digits 1 through 6 if the middle digit must be odd and no two digits might be the same?
A) 420
B) 360
C) 180
D) 120
E) 60
wHY THE ANSWER IS not 360 ?
3 * 5 * 4 * 3 * 2 ?
Thanks
five digit codes
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Hi yass20015,yass20015 wrote:How many different five-digit codes can be picked from the digits 1 through 6 if the middle digit must be odd and no two digits might be the same?
A) 420
B) 360
C) 180
D) 120
E) 60
wHY THE ANSWER IS not 360 ?
3 * 5 * 4 * 3 * 2 ?
Thanks
your equation is correct and 360 is the answer.
Ways to choose middle digit = 3 [1,3,5]
Ways to choose 1st digit = 5 [total digits - 1]
Ways to choose 2nd digit = 4 [total digits - 1-1]
Ways to choose 4th digit = 3 [total digits - 1-1-1]
Ways to choose 5th digit = 2 [total digits - 1-1-1-1]
total ways = 3x5x4x3x2=360
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To confirm, 360 is correct. If the answer is different, then question is incorrect. With problems like this, always start with your biggest constraint --> the middle number must be odd.
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Always post the source and the OA when posting a problem. Not only is that legally required, but it helps students to know which resources are good, and which contain faulty problems.
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Take the task of "building" 5-digit numnbers and break it into stages.yass20015 wrote:How many different five-digit codes can be picked from the digits 1 through 6 if the middle digit must be odd and no two digits might be the same?
A) 420
B) 360
C) 180
D) 120
E) 60
We'll begin with the most restrictive stage.
Stage 1: Select the MIDDLE (hundreds) digit
This digit must be odd (1, 3, or 5)
So, we can complete stage 1 in 3 ways
Stage 2: Select the units digit
There are 5 remaining digits from which to choose, so we can complete this stage in 5 ways.
Stage 3: Select the tens digit
There are 4 remaining digits from which to choose, so we can complete this stage in 4 ways.
Stage 4: Select the thousands digit
We can complete this stage in 3 ways.
Stage 5: Select the ten thousands digit
We can complete this stage in 2 ways.
By the Fundamental Counting Principle (FCP), we can complete all 5 stages (and thus build a 5-digit number) in (3)(5)(4)(3)(2) ways ([spoiler]= 360 ways[/spoiler])
Answer: B
--------------------------
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Ignoring the part about the middle digit having to be odd, there are 6x5x4x3x2 codesyass20015 wrote:How many different five-digit codes can be picked from the digits 1 through 6 if the middle digit must be odd and no two digits might be the same?
A) 420
B) 360
C) 180
D) 120
E) 60
Half of those will have an odd number in the middle, so the answer is (6x5x4x3x2)/2 = 360