Aforward path from (0,0) to (n, n) is good if it never goes strictly above the diagonal line x = y. any other forward path is bad. from class, the number of good forward paths is the nth catalan number.

in this problem, you will get a new derivation for the formula for catalan numbers without using generating functions.
we denote paths as sequences (v1, v2n) where each vi is either the vector (1,0) or (0,1).

(a) given a bad path ( v2n) from (0,0) to (n, n), let r be the smallest index such that vi + vr is above the line x = y, i. e., the second coordinate is strictly bigger than the first coordinate.
create a new path (w1, w2n) by w ; jvi if 1 < i show that w is a forward path from (0,0) to (n â 1, n + 1).

(b) in (a) we defined a function {bad forward paths from (0,0) to (n, n)} + {forward paths from (0,0) to (n â 1, n + 1)}. show that this function is a bijection.

mx+89b(-98)nb

step-by-step explanation:

if gloria has "carry over" minutes, (that is, any minutes paid for but not used in a month can be applied to her payment for minutes in the next month,) her total usage for the three months is: total minutes = 320 + 243 + 489 = 1052 minutes number of 200 minute blocks = 1052/200 = 5 r52 (that is 5 with a remainder of 52) rounding up the number of 200 minute blocks gives her usage for the three months as 6 blocks. the cost for six blocks is 6 x $25 = $150 she also has 200 - 52 or 148 carry over minutes if gloria pays each month and unused minutes are lost, applying the rounding up procedure to each month gives: first month usage of 320 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 second month usage of 243 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 third month usage of 489 minutes rounds up to 3 blocks of 200 minutes for a payment of 3 x $25 = $75 total payments for the three months are $50 + $50 + $75 = $175 get the plan with carry over minutes!