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Question Number 122954 by mr W last updated on 21/Nov/20

Commented by Dwaipayan Shikari last updated on 21/Nov/20

$I n f i n i t e l y m a n y a n s w e r s s i r!$ $I f t h e y a r e i n a r e l a t i o n C = 1 + (n - 1)$ $t h e n 12 w i l l c o s t 12 $ (f o r t h e p e o p l e w h o . . . . .)$ $I f t h e y a r e i n a r e l a t i o n$ $n + (n - 1) (n - 2) (n - 3) . . . (n - 11)$ $I f n = 12$ $t h e n p r i c e w i l l b e 12 + 11! $ :)$ $. . . . . .$
Commented by prakash jain last updated on 22/Nov/20

$Information about 2 and 11 is$ $redundant .$ $12 = 12 times ONE$ $So 12 will cost $ 12 .$
Commented by mr W last updated on 22/Nov/20

$a n o t h e r p o s s i b i l i t y :$ $s a y a l e t t e r O c o s t s o $, a l e t t e r N$ $c o s t s n $, e t c .$ $o + n + e = 1 . . . (i)$ $t + w + o = 2 . . . (i i)$ $3 e + l + v + n = 11 . . . (i i i)$ $(i i i) + (i i) - (i) :$ $2 e + l + v + t + w = 12 . . . (i v)$ $T W E L V E c o s t s 2 e + t + w + l + v, t h a t$ $i s 12 $ f r o m (i v) .$
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