Mistakes give rise to Problems- 14

By the converse of theorem on equal ratios, we conclude that $\dfrac{a_i}{b_i} = k \quad \quad \forall i \in (1,2,3,4)$

Hence we want to find sequences of length 4 of < $a_i$ > and < $b_i$ such that $\dfrac{a_1}{b_1}= \dfrac{a_2}{b_2}=\dfrac{a_3}{b_3}=\dfrac{a_4}{b_4}$ and are integers in the range [1,20].

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We know that only numbers which have at least 4 multiples in the range [1,20] are $1,2,3,4$ and $5$ .

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Hence we will have $19$ different fractions possible (fractions are the values of $k$ possible), namely $1,\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},\dfrac{1}{5},\dfrac{2}{3},\dfrac{2}{5},\dfrac{3}{4},\dfrac{3}{5},\dfrac{4}{5}$ and their reciprocals. (only 1 won't give a different reciprocal and 2/4 is same as 1/2).

Thus the equal ratios $\dfrac{a_i}{b_i}$ can have 19 distinct values ( $k$ )

We'll find the number of sequences with the greater of the two numbers in the fraction because the smaller number will adjust it's multiples to keep the value same.

And that will always be $\dbinom{q}{4}$ where $q$ is defined as number of multiples less than 21, of the number maximum out of $x$ and $y$ where $x$ and $y$ are numbers included in the fraction.

Here, $5$ will be the maximum number in $8$ fractions, namely $\dfrac{1}{5},\dfrac{2}{5},\dfrac{3}{5},\dfrac{4}{5}$ and their reciprocals .

Thus, the number sequences for this will be $8\times \dbinom{4}{4}$

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$4$ will be the maximum number in $4$ fractions, namely $\dfrac{1}{4} , \dfrac{3}{4}$ and their reciprocals.

Thus $4\times \dbinom{5}{4}$ sequences pairs.

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$3$ will be the maximum number in $4$ fractions, namely $\dfrac{1}{3} , \dfrac{2}{3}$ and their reciprocals.

Thus $4\times \dbinom{6}{4}$ pairs of sequences.

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$2$ will be maximum number in $2$ fractions, which are $\dfrac{1}{2}$ and $\dfrac{2}{1}$

Thus $2\times \dbinom{10}{4}$ pairs of sequences.

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$1$ will be maximum in only $\dfrac{1}{1}$ ,which gives $\dbinom{20}{4}$ pairs of sequences.

Hence there will be total of $\dbinom{20}{4}+2\times \dbinom{10}{4}+4\times \dbinom{6}{4}+4\times \dbinom{5}{4}+8\times \dbinom{4}{4} = \boxed{5353}$