Compute the number of ordered quadruple (a, b, c, d) of distinct positive integers so that (( ((a),(b) )),( ((c),(d) )) ) = 21 .


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Question Number 32569 by naka3546 last updated on 28/Mar/18

$C o m p u t e t h e n u m b e r o f o r d e r e d q u a d r u p l e (a, b, c, d) o f d i s t i n c t p o s i t i v e i n t e g e r s s o t h a t (\begin{matrix} (\begin{matrix} a \\ b \end{matrix}) \\ (\begin{matrix} c \\ d \end{matrix}) \end{matrix}) = 21 .$
Commented by MJS last updated on 28/Mar/18

$(\begin{matrix} n \\ k \end{matrix}) = 21 \Rightarrow (n; k) \in {(7; 2); (7; 5); (21; 1); (21; 20)}$ $(\begin{matrix} n \\ k \end{matrix}) = 1 \Rightarrow (n; k) = (m; m) ∣ m \in N$ $but m = m is not allowed, and$ $0 {isn}^{'} t allowed too$ $(\begin{matrix} n \\ k \end{matrix}) = 2 \Rightarrow (n; k) = (2; 1)$ $(\begin{matrix} n \\ k \end{matrix}) = 5 \Rightarrow (n; k) \in {(5; 1); (5; 4)}$ $(\begin{matrix} n \\ k \end{matrix}) = 7 \Rightarrow (n; k) \in {(7; 1); (7; 6)}$ $(\begin{matrix} n \\ k \end{matrix}) = 20 \Rightarrow (n; k) \in {(6; 3); (20; 1); (20; 19)}$ $(\begin{matrix} n \\ k \end{matrix}) = 21 \Rightarrow (n; k) \in {(7; 2); (7; 5); (21; 1); (21; 20)}$ $so there are 13 quadruples (a; b; c; d) :$ $(7; 6; 2; 1)$ $(7; 1; 5; 4)$ $(7; 6; 5; 1)$ $(7; 6; 5; 4)$ $(7; 2; 6; 3)$ $(7; 2; 20; 1)$ $(7; 2; 20; 19)$ $(7; 5; 6; 3)$ $(7; 5; 20; 1)$ $(7; 5; 20; 19)$ $(21; 1; 6; 3)$ $(21; 1; 20; 19)$ $(21; 20; 6; 3)$
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