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Find the numbers between $1500$ and $5000$ that can be made from the digits $1,2,4,5,7$ and $8$ if each digit is only used once.

Can someone show the steps?

N. F. Taussig
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Heng Yong Wei
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1 Answers1

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You have to add up a few cases:

  1. 15AB
  2. 1ABC (where A = 7 or A = 8)
  3. ABCD (where A = 2 or A = 4)

And per case count how many ways you can select (valid) combination.

It's just manual labor.

(Given that 1 and 2 can be combined:

  1. $1ABC$, where $A \in \{5,7,8\}$ and select 2 from $\{2,4,a,b\}$ gives $1*3*4*3 = 36$ numbers. $a,b$ are the leftovers after selecting $A$ from $5,7,8$.

  2. $ABCD$, where $A \in \{2, 4\}$ and select 3 from $\{1,5,7,8,c\}$ gives $2*5*4*3 = 120$ numbers

In total: $36 + 120 = 156$.

Pieter21
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    Your case 1 and case 2 can really be put into a single case. – Arthur Nov 11 '16 at 10:26
  • @Arthur, indeed. Good optimization that is valid in this case. I leave my answer as is though, it is a bit more fool proof method for similar problems, – Pieter21 Nov 11 '16 at 10:33
  • I see. But how to find the answer using calculator? – Heng Yong Wei Nov 11 '16 at 11:37
  • You haven't accounted for permutations between the numbers. The answer to this question must be 156. In the ABCD case, after you have selected A, the remaining three letters can also permute amongst themselves in 3! ways. Answer should be 12(1st case)+24(2nd case)+120(3rd case)=156 – Arishta May 27 '17 at 05:56
  • @Cotton, you're right! I'll modify the answer.. – Pieter21 May 27 '17 at 13:01