Pretend you're playing a carnival game and you've won the lottery, sort of. You have the opportunity to select five bills from a money bag, while blindfolded. The bill values are $1, $2, $5, $10, $20, $50, and $100. How many different possible ways can you choose the five bills? (Order doesn't matter, and there are at least five of each type of bill.) A. 56 B. 120 C. 288 D. 462

Accepted Solution

Answer:The total number of ways are:                        462Step-by-step explanation:When we are asked to select r items from a set of n items that the rule that is used to solve the problem is:Method of combination.Here the total number of bills of different values are: 7i.e. n=7(  $1, $2, $5, $10, $20, $50, and $100 )and there are atleast five of each type of bill.Also, we have to choose 5 bills i.e. r=5The repetition is  allowed while choosing bills.Hence, the formula is given by:[tex]C(n+r-1,r)[/tex]Hence, we get:[tex]C(7+5-1,5)\\\\i.e.\\\\C(11,5)=\dfrac{11!}{5!\times (11-5)!}\\\\C(11,5)=\dfrac{11!}{5!\times 6!}\\\\\\C(11,5)=\dfrac{11\times 10\times 9\times 8\times 7\times 6!}{5!\times 6!}\\\\\\C(11,5)=\dfrac{11\times 10\times 9\times 8\times 7}{5!}\\\\\\C(11,5)=\dfrac{11\times 10\times 9\times 8\times 7}{5\times 4\times 3\times 2}\\\\\\C(11,5)=462[/tex]            Hence, the answer is:                   462