Compute-the-number-of-ordered-quadruple-a-b-c-d-of-distinct-positive-integers-so-that-a-b-c-d-21-

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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