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Permutation /Combination

by mac007 » Fri Dec 12, 2014 3:32 am
There is 4 coins 1 paisa, 5 paise, 10 paise, 25 paise using these coins we have to make 50 paisa how many combination can we make ?
please explain your apprach in detail...

Here is my solution..........please tell me if this is correct.....can someone solve this using some simple-short method ?

p+5q+10r+25s=50
we'll take cases for s=0,1 and 2
for s=0,r=0,q can be 0 to 10=11 combinations
for s=0,r=1,q can be 0 to 8=9 combinations
for s=0,r=2=>q=0 to 6=7 combinations
for s=0,r=3=>q=0-4=5 combinations
for s=0,r=4=>q=0-2=3 com
s=0,r=5=>q=0=1 com
for s=1,r=0=>q=0-5=6 comb
s=1,r=1==>q=0-3=4 comb
s=1,r=2 ==>q=0-1=2 comb
s=2,r=0,q=0,p=0==>1 comb
hence total combinations are 49
Note:there p's condition is already considered when we are taking values for q.
=> && ==> means not greater this is used as i.e...
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by [email protected] » Sat Dec 13, 2014 12:39 pm
Hi mac007,

Did this question come from a GMAT source (book, Course, etc.)? If so, then you'd have the benefit of 5 answer choices to look at (and possibly avoid some of this work). Before I offer a detailed solution, I'd like to have the entire question to work with.

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by Matt@VeritasPrep » Mon Dec 22, 2014 2:49 pm
The problem essentially reads
Mac has x pennies, y nickels, z dimes, and w quarters. The sum of the values of his coins is 50 cents. If x, y, z, and w are positive integers, how many distinct sets {x,y,z,w} are possible?
or, in a friendlier way,
How many different ways can we make 50¢ from a bag full of 50 pennies, 10 nickels, 5 dimes, and 2 quarters, if we can remove as many of each coin as we like from the bag?
I wish I could think of a way that doesn't involve casework, but that's all I've been able to find. Here's a neat approach though.

If we use two quarters there's obviously only one way to solve this.

If we use one quarter, we'll have a few more possibilities. A nice way to do this is to consider the number of pennies used. If we use 25 pennies, we're done. If we use 20 pennies, we're also done: the remaining five cents must be a nickel. If we use 15 pennies, however, we have two options: a dime OR two nickels. If we use 10, same idea: two options, a dime and a nickel OR three nickels.

You can probably guess where this is going. If we use 5 pennies, we'll have three possibilities. If we use 0 pennies, we'll also have three possibilities. At this point, we're done.

This gives us 1 + 1 + 2 + 2 + 3 + 3 ways of grouping with one quarter.

If we use no quarters, we can follow the same logic. With 50 pennies, we can one way. With 45, we also have one way. But 40 and 35 each give us two ways, 30 and 25 each give us three ways, etc. We'll thus have 1 + 1 + 2 + 2 + 3 + 3 + 4 + 4 + 5 + 5 + 6 ways.

This gives us
Two quarters = 1 way
One quarter = 1+1+2+2+3+3 = 12 ways
No quarters = 1+1+2+2+3+3+4+4+5+5+6 = 36 ways

for a total of 49 ways of making 50¢ with our bag of coins.

Note that this generalization works for ANY sum to be made with these four coins. For instance, if we had to make a dollar (100 cents) with these coins, we'd have

Four quarters = 1 way
Three quarters = 1+1+2+2+3+3 = 12 ways
Two quarters = 1+1+2+2+3+3+4+4+5+5+6 = 36 ways
One quarter = 1+1+2+2+...+8+8 = 72 ways
No quarters = 1+1+2+2+...+10+10+11 = 121 ways

and of course you can generalize further from here.
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