Question

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.)

Answer 1

Answer:

The total number of ways are:

                       462

Step-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: 7

i.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=5

The repetition is  allowed while choosing bills.

Hence, the formula is given by:

$C(n+r-1,r)$

Hence, we get:

$C(7+5-1,5)\\\\i.e.\\\\C(11,5)=\dfrac{11!}{5!\times (11-5)!}\\\\C(11,5)=\dfrac{11!}{5!\times 6!}\\\\\\C(11,5)=\dfrac{11\times 10\times 9\times 8\times 7\times 6!}{5!\times 6!}\\\\\\C(11,5)=\dfrac{11\times 10\times 9\times 8\times 7}{5!}\\\\\\C(11,5)=\dfrac{11\times 10\times 9\times 8\times 7}{5\times 4\times 3\times 2}\\\\\\C(11,5)=462$

           Hence, the answer is:

                  462