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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.)

User Baskren
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1 Answer

1 vote

Answer:

The total number of ways are:

462

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)=(11!)/(5!* (11-5)!)\\\\C(11,5)=(11!)/(5!* 6!)\\\\\\C(11,5)=(11* 10* 9* 8* 7* 6!)/(5!* 6!)\\\\\\C(11,5)=(11* 10* 9* 8* 7)/(5!)\\\\\\C(11,5)=(11* 10* 9* 8* 7)/(5* 4* 3* 2)\\\\\\C(11,5)=462

Hence, the answer is:

462

User Beier
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5.0k points