How many sets of three integers between 1 and 20 are possible if no two consecutive integers are to be a set?

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20-01-2017
Mathematics

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How many sets of three integers between 1 and 20 are possible if no two consecutive integers are to be a set?

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LammettHash LammettHash · Answer 1 · 2017-01-20T17:29:53+01:00

There are [tex]\dbinom{20}3=1140[/tex] total possible ways to pick any three integers from the set.

Of the total, there are 18 consisting of consecutive triplets [tex](\{1,2,3\},\{2,3,4\},\ldots,\{18,19,20\})[/tex].

Now, of the total, suppose you fix two integers to be consecutive. There would be 19 possible pairs [tex](\{1,2\},\{2,3\},\ldots,\{19,20\})[/tex], and for each pair 18 possible choices for the third integer (for instance, [tex]\{1,2\}[/tex] can be taken with 3, 4, ..., 20), to a total of [tex]19\times18=342[/tex]. To avoid double-counting (e.g. [tex]\{1,2\}[/tex] can't go with 3; [tex]\{2,3\}[/tex] can't go with 1 or 4), we subtract 1 from the extreme pairs [tex]\{1,2\}[/tex] and [tex]\{19,20\}[/tex] (twice), and 2 from the rest (17 times).

So, the number of triplets that don't consist of pairwise consecutive integers is

[tex]1140-(18+342-(2\times1+17\times2))=816[/tex]

I don't know how useful this would be to you, but I've verified the count in Mathematica:

In[8]:= DeleteCases[Subsets[Range[1, 20], {3}], x_ /; x[[2]] == x[[1]] + 1 || x[[3]] == x[[2]] + 1] // Length
Out[8]= 816

MrRoyal MrRoyal · Answer 2 · 2021-10-08T06:15:50+02:00

The question is an illustration of combination (i.e. selections)

There are 816 sets of three non-consecutive integers between 1 and 20

The given parameters are:

[tex]\mathbf{N = 20}[/tex] --- number of integers

[tex]\mathbf{r = 3}[/tex] --- the set of integers

For the integers to be selected such that there are no two consecutive integers.

The value of n must be:

[tex]\mathbf{n = N - r + 1}[/tex]

So, we have:

[tex]\mathbf{n = 20 - 3 + 1}[/tex]

[tex]\mathbf{n = 18}[/tex]

The number of selection is then calculated as:

[tex]\mathbf{^nC_r = \frac{n!}{(n - r)!r!}}[/tex]

So, we have:

[tex]\mathbf{^{18}C_3 = \frac{18!}{(18 - 3)!3!}}[/tex]

[tex]\mathbf{^{18}C_3 = \frac{18!}{15!3!}}[/tex]

Expand

[tex]\mathbf{^{18}C_3 = \frac{18 \times 17 \times 16 \times 15!}{15! \times 3 \times 2 \times 1}}[/tex]

[tex]\mathbf{^{18}C_3 = \frac{18 \times 17 \times 16}{3 \times 2 \times 1}}[/tex]

[tex]\mathbf{^{18}C_3 =3 \times 17 \times 16}[/tex]

[tex]\mathbf{^{18}C_3 =816}[/tex]

Hence, there are 816 sets of three non-consecutive integers between 1 and 20

Read more about combinations (selection) at:

https://brainly.com/question/4546043