There are 4 appetizers total. We want to select 3 of them. Order doesn't matter. Use the nCr formula with n = 4 and r = 3 n C r = (n!)/(r!*(n-r)!) 4 C 3 = (4!)/(3!*(4-3)!) 4 C 3 = (4!)/(3!*1!) 4 C 3 = (4*3!)/(3!*1!) 4 C 3 = (4)/(1!) 4 C 3 = (4)/(1) 4 C 3 = 4/1 4 C 3 = 4 There are 4 ways to select just the appetizers Call this value M = 4 (we'll use it later) ---------------------------- There are 10 main courses total. We want to select 8 of them. Order doesn't matter. Use the nCr formula with n = 10 and r = 8 n C r = (n!)/(r!*(n-r)!) 10 C 8 = (10!)/(8!*(10-8)!) 10 C 8 = (10!)/(8!*2!) 10 C 8 = (10*9*8!)/(8!*2!) 10 C 8 = (10*9)/(2!) 10 C 8 = (10*9)/(2*1) 10 C 8 = 90/2 10 C 8 = 45 There are 45 ways to select just the main courses Call this value N = 45 (we'll use it later) ---------------------------- There are 9 desserts total. We want to select 3 of them. Order doesn't matter. Use the nCr formula with n =9 and r = 3 n C r = (n!)/(r!*(n-r)!) 9 C 3 = (9!)/(3!*(9-3)!) 9 C 3 = (9!)/(3!*6!) 9 C 3 = (9*8*7*6!)/(3!*6!) 9 C 3 = (9*8*7)/(3!) 9 C 3 = (9*8*7)/(3*2*1) 9 C 3 = 504/6 9 C 3 = 84 There are 84 ways to select just the desserts Call this value P = 84 (we'll use it later) ---------------------------- Multiply the results from earlier: M*N*P = 4*45*84 = 15120 which is the final result.
Step-by-step explanation: because of the decimal you have to place a zero regardless and the other 2 zeros you can’t do anything with them. Hope this helps!
