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How many 7-digit numbers exists such that no digits are repeated and 8 and 9 are not adjacent? (If 8 and/or 9 don't appear in the number that counts as not being adjacent)

(A)181400   (B)362880   (C)436800   (D)472080   (E)544320

 

I attempted to do casework on this but I was not able to successfully wrangle out an answer; can someone help me with this problem? Thanks

 Apr 4, 2020
 #1
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I'm getting an answer choice (D), yet I might be wrong.

 

Instead of using casework, it's better to use complementary counting and caseowrk.

 

To start off, the number of ways to from seven-digit numbers such that no digits are repeating are: 9*9*8*7*6*5*4=544320 ways(numbers).

 

Next, imagine we have the following formation:

 

 8   9   _   _  _  _  _ 

 

For the first blank, I have eight choices (excluding eight and nine, including zero)...then seven choices for the second blank, six for the third blank, and so on for 8*7*6*5*4=6720 numbers.

 

Note that the eight and nine are can be interchanged, so we have 2 * 6720=13440 ways(numbers).

 

 

Next, imagine eight and nine are in the second and third positions respectfully. 

 

_ 8 9 _ _ _ _

 

There are seven choices for the first digit(excluding zero), seven for the fourth digit(including zero), six choices for the fourth digit and so on.

 

Thus, we have 7 * 7 * 6 * 5 * 4= 5880 cases.

 

This same case can happen in positions three and four, four and five, five and six, and finally six and seven.

 

5880 * 2 * 5=58800 ways(numbers).

 

Yes, thus, we have 544320-13440-58800=472080 ways(numbers) for choice (D).

 

Again, I might be wrong.

 Apr 4, 2020
 #2
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Thank you so much!

Guest Apr 4, 2020

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