SOLUTION: How many 5 digit PINs (using digits 0-9) are possible if numbers can be repeated, but the repeated numbers must be odd?

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Question 971308: How many 5 digit PINs (using digits 0-9) are possible if numbers can be repeated,
but the repeated numbers must be odd?

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
How many 5 digit PINs (using digits 0-9) are possible if numbers can be
repeated, but the repeated numbers must be odd?
There are 5 even digits and 5 odd digits.

I'll go through it step-by-step, case-by-case, below, but we can write it in
summation form:



For k = 0 to 5
Choose k evens 5Ck ways 
Position the k evens in the PIN 5Pk ways
Choose any of 5 odd digits for each of the remaining 5-k positions.

k= 0 through 5 amounts to 6 cases:

1. 0 evens, 5 odds

Choose the 5 evens 5^5 ways. That's 3125 ways.

So there are 3125 = 5C0*5P0*5^5 ways for case 1.

2. 1 even, 4 odds

Choose the even 5C1 ways. That's 5 ways.
Choose the position for the even 5C1 ways. That's 5 ways.
Choose the 4 odds 5^4 ways. That's 625 ways.

So there are 5*5*625 = 5C1*5P1*5^4 = 15625 ways for case 2.

3. 2 evens, 3 odds

Choose the evens 5C2 ways. That's 10 ways.
Place the smaller even in the PIN 5 ways.
Place the larger even in the PIN 4 ways.
Choose the 3 odds in the 3 remaining positions 5^3 ways. That's 125 ways.

So there are 10*5*4*125 = 5C2*5P2*5^3 = 25000 ways for case 3.

4. 3 evens, 2 odds

Choose the evens 5C3 ways. That's 10 ways.
Place the smallest even in the PIN 5 ways.
Place the middle-sized even in the PIN 4 ways.
Place the largest even in the PIN 3 ways.
Choose the 2 odds in the 2 remaining positions 5^2 ways. That's 25 ways.

So there are 10*5*4*3*25 = 5C3*5P3*5^2 = 15000 ways for case 4.

5. 4 evens, 1 odd

Choose the evens 5C4 ways. That's 5 ways.
Place the smallest even in the PIN 5 ways.
Place the next to smallest even in the PIN 4 ways.
Place the next to largest even in the PIN 3 ways.
Place the largest even in the PIN 2 ways.
Choose the 1 odd in the 1 remaining position 5C1 ways. That's 5 ways.

So there are 5*5*4*3*2*5 = 5C4*5P4*5^1 = 3000 ways for case 5.

6. 5 evens, 0 odds

Place the 0 in the PIN 5 ways.
Place the 2 in the PIN 4 ways.
Place the 4 in the PIN 3 ways.
Place the 6 in the PIN 2 ways.
Place the 8 in the PIN 1 way.

That's 5! = 5P5 = 5*4*3*2*1 = 120 for case 6.

That's 5C5*5P5*5^0 = 120 ways for case 6


Grand total:

 3125 for case 1
15625 for case 2
25000 for case 3
15000 for case 4
 3000 for case 5
  120 for case 6
----------------
61870 ways total



Edwin