How many 6-digit numbers?

Probability Level 3

If we use 0 0 , 1 1 , 2 2 , 3 3 , 4 4 , and 5 5 , each of these integers only once, to make a valid 6-digit number, such that even integers are not next to each other and 0 0 and 5 5 must be next to each other, how many such 6-digit numbers can we make?


The answer is 60.

Alice Smith by Alice Smith , 20, China

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1 solution

David Vreken
Feb 19, 2020

There are four ways even integers can be arranged in a 6 6 -digit number so that they are not next to each other (e for even and o for odd): eoeoeo, oeoeoe, eoeooe, and eooeoe. To be a valid 6 6 -digit number, it cannot start with 0 0 . The possibilities are then:

eoeoeo → 2 × 3 × 2 × 2 × 1 × 1 = 24 \text{eoeoeo} \rightarrow 2 \times 3 \times 2 \times 2 \times 1 \times 1 = 24

oeoeoe → 3 × 3 × 2 × 2 × 1 × 1 = 36 \text{oeoeoe} \rightarrow 3 \times 3 \times 2 \times 2 \times 1 \times 1 = 36

eoeooe → 2 × 3 × 2 × 2 × 1 × 1 = 24 \text{eoeooe} \rightarrow 2 \times 3 \times 2 \times 2 \times 1 \times 1 = 24

eooeoe → 2 × 3 × 2 × 2 × 1 × 1 = 24 \text{eooeoe} \rightarrow 2 \times 3 \times 2 \times 2 \times 1 \times 1 = 24

For a total of 24 + 36 + 24 + 24 = 108 24 + 36 + 24 + 24 = 108 possibilities.

However, since 0 0 and 5 5 must be next to each other, we must also subtract out the possibilities where 0 0 and 5 5 are not next to each other. The possibilities are:

eoeoeo: \text{eoeoeo:}

  • eo0oe5 → 2 × 2 × 1 × 1 × 1 × 1 = 4 \text{eo0oe5} \rightarrow 2 \times 2 \times 1 \times 1 \times 1 \times 1 = 4
  • e5eo0o → 2 × 1 × 1 × 2 × 1 × 1 = 4 \text{e5eo0o} \rightarrow 2 \times 1 \times 1 \times 2 \times 1 \times 1 = 4

oeoeoe: \text{oeoeoe:}

  • o0oe5e → 2 × 1 × 1 × 2 × 1 × 1 = 4 \text{o0oe5e} \rightarrow 2 \times 1 \times 1 \times 2 \times 1 \times 1 = 4
  • 5eo0oe → 1 × 2 × 2 × 1 × 1 × 1 = 4 \text{5eo0oe} \rightarrow 1 \times 2 \times 2 \times 1 \times 1 \times 1 = 4
  • 5eoeo0 → 1 × 2 × 2 × 1 × 1 × 1 = 4 \text{5eoeo0} \rightarrow 1 \times 2 \times 2 \times 1 \times 1 \times 1 = 4
  • oe5eo0 → 2 × 2 × 1 × 1 × 1 × 1 = 4 \text{oe5eo0} \rightarrow 2 \times 2 \times 1 \times 1 \times 1 \times 1 = 4

eoeooe: \text{eoeooe:}

  • eo0o5e → 2 × 2 × 1 × 1 × 1 × 1 = 4 \text{eo0o5e} \rightarrow 2 \times 2 \times 1 \times 1 \times 1 \times 1 = 4
  • e5eoo0 → 2 × 1 × 1 × 2 × 1 × 1 = 4 \text{e5eoo0} \rightarrow 2 \times 1 \times 1 \times 2 \times 1 \times 1 = 4
  • eoe5o0 → 2 × 2 × 1 × 1 × 1 × 1 = 4 \text{eoe5o0} \rightarrow 2 \times 2 \times 1 \times 1 \times 1 \times 1 = 4

eooeoe: \text{eooeoe:}

  • e5o0oe → 2 × 1 × 2 × 1 × 1 × 1 = 4 \text{e5o0oe} \rightarrow 2 \times 1 \times 2 \times 1 \times 1 \times 1 = 4
  • e5oeo0 → 2 × 1 × 2 × 1 × 1 × 1 = 4 \text{e5oeo0} \rightarrow 2 \times 1 \times 2 \times 1 \times 1 \times 1 = 4
  • eo5eo0 → 2 × 2 × 1 × 1 × 1 × 1 = 4 \text{eo5eo0} \rightarrow 2 \times 2 \times 1 \times 1 \times 1 \times 1 = 4

For a total of 12 ⋅ 4 = 48 12 \cdot 4 = 48 possible ways where 0 0 and 5 5 are not next to each other.

Therefore, with the given restrictions there are 108 − 48 = 60 108 - 48 = \boxed{60} possible 6 6 -digit numbers.

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