Verified Answer :

LammettHash

There are \(\dbinom{20}3=1140\) total possible ways to pick any three integers from the set.

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

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

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

\(1140-(18+342-(2\times1+17\times2))=816\)

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

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:

\(\mathbf{N = 20}\) --- number of integers

\(\mathbf{r = 3}\) --- the set of integers

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

The value of n must be:

\(\mathbf{n = N - r + 1}\)

So, we have:

\(\mathbf{n = 20 - 3 + 1}\)

\(\mathbf{n = 18}\)

The number of selection is then calculated as:

\(\mathbf{^nC_r = \frac{n!}{(n - r)!r!}}\)

So, we have:

\(\mathbf{^{18}C_3 = \frac{18!}{(18 - 3)!3!}}\)

\(\mathbf{^{18}C_3 = \frac{18!}{15!3!}}\)

Expand

\(\mathbf{^{18}C_3 = \frac{18 \times 17 \times 16 \times 15!}{15! \times 3 \times 2 \times 1}}\)

\(\mathbf{^{18}C_3 = \frac{18 \times 17 \times 16}{3 \times 2 \times 1}}\)

\(\mathbf{^{18}C_3 =3 \times 17 \times 16}\)

\(\mathbf{^{18}C_3 =816}\)

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

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