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# Bill purchases an item and receives no change. Before the purchase, he

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Joined: 28 Apr 2013
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Bill purchases an item and receives no change. Before the purchase, he  [#permalink]

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Updated on: 05 Jun 2015, 04:22
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Difficulty:

55% (hard)

Question Stats:

61% (01:30) correct 39% (01:45) wrong based on 101 sessions

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Bill purchases an item and receives no change. Before the purchase, he has only a five dollar bill, two 10 dollar bill, and a twenty dollar bill. How many distinct possibilities are there for the total amount of his purchase?

A. 3
B. 4
C. 6
D. 9
E. 10

Can anybody solve this question using permutation combination method, if it is possible to do so.

Originally posted by mpingo on 04 Jun 2015, 20:28.
Last edited by Bunuel on 05 Jun 2015, 04:22, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Bill purchases an item and receives no change. Before the purchase, he  [#permalink]

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04 Jun 2015, 21:59
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Hi mpingo,

Using permutations/combinations here would be the "long" method. This question has a great 'brute force' approach that you can use to answer it relatively quickly. You can tell from the answer choices that there are AT LEAST 3 possible totals, but NO MORE THAN 10 possible totals, so you just need to list them out.

We have 5, 10, 10 and 20 to work with

From least to greatest, we have...
5 (the least)
10
15
20
25
30
35
40
45 (the greatest)

Total options = 9

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Re: Bill purchases an item and receives no change. Before the purchase, he  [#permalink]

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04 Jun 2015, 22:04
1
mpingo wrote:
Can anybody solve this question using permutation combination method, if it is possible to do so.

Bill purchases an item and receives no change. Before the purchase, he has only a five dollar bill, two 10 dollar bill, and a twenty dollar bill. How many distinct possibilities are there for the total amount of his purchase?

a.3
b.4
c.6
d.9
e.10

You can solve it using Combinatorics though I think the easiest method here would be to work on distinct amounts.
Just quickly check for multiples of 5 from 5 to 45 - You can make only multiples of 5 with 5, 10 and 20 and the maximum you can make is 5 + 10 + 10 + 20 = 45
So imagine 4 currency bills: 5 ,10, 10, 20
Now, can you make 10? Yes; 15? Yes; 20 Yes... till 45. You can make all these amounts and hence there are 9 distinct amounts.
The reason this method is the best is that you work on distinct amounts, not on distinct bills. So there is no worry of getting the same amount in two cases and not realising it.

A Combinatorics Method:

Considering the 3 distinct bills:
Amounts made from 1 bill: 3C1 ways
5, 10, 20

Amounts made from 2 bills: 3C2 (any two out of 5, 10, 20) + 1 (10 and 10)
Amounts made: 15, 25, 30, 20
Total 6 distinct amounts

Amounts made from 3 bills: 3C3 + 2 (two 10s and one other)

Amounts made from all 4 bills: 5+10+10+20 = 45

Total 9 distinct amounts.
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Re: Bill purchases an item and receives no change. Before the purchase, he  [#permalink]

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23 Jun 2015, 06:25
I wrote down the possibilities we could have:

Five: 1
Ten: 2
Twenty: 1

So, starting low, ie, into the number of possibilities each bill alone is giving:
Five gives 1 possibility (1), Ten gives 2 possibilities (10 or 20) and Twenty gives 1 possibility (20).

Combining Five and Ten:
We have 2 possibilities (15 or 25).

Combining Five and Twenty:
We have 1 possibility (25)

Combining Ten and Twenty:
We have 2 possibilities (30 or 20).

Counting all the possibilities we have in total 9 possible amounts that could have been paid. ANS D
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Re: Bill purchases an item and receives no change. Before the purchase, he  [#permalink]

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04 Oct 2017, 06:17
We have one 5 \$ bill, two 10 \$ bills and one 20 \$ bill.
1) We can have only 5\$
2) 5+10= 15\$
3) 5+10+10= 25\$
4) 10\$
5) 10+10= 20\$
a) 20\$--- redundant
b) 20+5=25\$--redundant do not count
6) 20+10= 30\$
7) 20+10+5= 35\$
8) 20+10+10= 40\$
9) 20+10+10+5= 45\$

Kudos if it helps.
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Re: Bill purchases an item and receives no change. Before the purchase, he  [#permalink]

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04 Oct 2017, 07:48
\$5 bill = 1; \$10 bills = 2; \$20 bill = 1. So total number of notes = 4.

Now, Bill can give either 1 note, or 2 notes, or 3 notes or 4 notes.

Case 1: Bill gave 1 Note
total number of outcomes = 4 i.e. \$5, \$10, \$10, \$20;

Case 2: Bill gave 2 Notes

total number of outcomes = 6 i.e. \$5 + \$10; \$5 + \$10; \$5 + \$20; \$10 + \$10; \$10 + \$20; and \$10 + \$20;

Case 3: Bill gave 3 Notes

total number of outcomes = 4 i.e. \$5+ \$10 + \$20; \$5+ \$10 + \$20; \$5+ \$10 + \$10; \$10+ \$10 + \$20;

Case 4: Bill gave 4 Notes

total number of outcomes = 1 i.e. \$5+ \$10+ \$10 + \$20;

So, total number of outcomes are 15 out of which unique outcomes are 9 i.e. \$5, \$10, \$15, \$20, \$25, \$30, \$35, \$40, \$45;

Ans: D
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