Mock test 5 - pending 21,23-28
Q1. (Combinatorics) "How many ordered quadruples (a, b, c, d) of positive odd integers are there that satisfy the equation a + b + c + 2d = 15?" Solution: Answer: 34 Why Rewrite every odd variable in terms of non-negative integers a = 2 x + 1 , b = 2 y + 1 , c = 2 z + 1 , d = 2 w + 1 , x , y , z , w ≥ 0 a = 2x+1,\; b = 2y+1,\; c = 2z+1,\; d = 2w+1,\qquad x,y,z,w \ge 0 Plug into the equation ( 2 x + 1 ) + ( 2 y + 1 ) + ( 2 z + 1 ) + 2 ( 2 w + 1 ) = 15 ⇒ 2 x + 2 y + 2 z + 4 w + 5 = 15 ⇒ 2 ( x + y + z + 2 w ) = 10 ⇒ x + y + z + 2 w = 5 (2x+1)+(2y+1)+(2z+1)+2(2w+1)=15 \Rightarrow 2x+2y+2z+4w+5 = 15 \Rightarrow 2(x+y+z+2w)=10 \Rightarrow x+y+z+2w = 5 Count solutions for x + y + z + 2 w = 5 x+y+z+2w=5 Because w w is an integer, list its possible values and use stars-and-bars for the remaining sum: w w equation for x + y + z x+y+z # solutions ( n + 2 2 ) \binom{n+2}{2} 0 x + y + z = 5 x+y+z = 5 ( 7 2 ) = 21 \binom{7}{2}=21 1 x + y ...