Combinations question- need help!!

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Combinations question- need help!!

by SM2010 » Tue Jun 29, 2010 1:13 pm
Hi... Even though this isn't a GMAT question it's a question that resembles the GMAT format, and it's taken from a Maths Competition taken by 16-19 year old high school students in the UK- It's called "Senior Mathematical Challenge" and it's good for those wanting to practice hard questions.

Question:
Given an unlimited supply of 50p, £1 and £2 coins, in how many different ways is it possible to make a sum of £100?

A) 1326
B) 2500
C) 2601
D) 5050
E) 10 000
Source: — Problem Solving |

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by Rich@VeritasPrep » Tue Jun 29, 2010 6:18 pm
Since the number of ways to choose 2-pound coins is fewer than the numbers of ways to do the other two coins, map out each possibility:

50 £2 coins - that already sums to £100 - 1 way

49 £2 coins - you could have 2 £1s; or you could have £1 and 2 50ps; or you could have0 £1 and 4 50ps - 3 ways

48 £2 coins - 4 £1s; 3 £1 and 2 50ps; 2 £1 and 4 50ps; 1 £1 and 6 50ps; 0 £1 and 8 50ps - 5 ways

47 £2 coins - 6 £1 coins ... ; 5 £1 coins ...; ... ... ... ; 0 £1 coins - 7 ways

You see the recurring pattern: You're progressing in consecutive odds. Continue the pattern:
...
...

1 £2 coin - 98 £1 coins; 97 £1 coins and 2 50ps; ... ... ... ; 0 £1 coin, - 99 ways

0 £2 coins - 100 £1 coins; 99 £1 coins and 2 50ps; ... ... ; 0 £1 coins and 200 50ps - 101 ways

Add up all the ways, and you get:

1+3+5+7+...+99+101

There are 51 numbers in that list. You can tell this because you're adding consecutive odd numbers. 1 = 2(0) + 1. 3 = 2(1)+1. And 101 = 2(50)+1. So you're adding all consecutive odds from 2(0)+1 to 2(50)+1. 51 terms.

You can group these into 50 pairs of numbers plus a 51 in the middle. Like so:

(101+1)+(99+3)+(97+5)+ ... + (53+49) + 51

That's 25 pairs of 102 plus the extra 51:

102 * 25 + 51 =

2601

You could also reach this number by identifying a pattern in sums of consecutive odds:

1 = 1
1+3 = 4
1+3+5 = 9
1+3+5+7 = 16

Notice that the sum of the first N consecutive positive odds is simply N^2.

Since we're adding the first 51 consecutive positive odds, our answer will simply be [spoiler]51^2 = 2601[/spoiler]
Rich Zwelling
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by SM2010 » Tue Jun 29, 2010 6:53 pm
That's the correct answer.... Thanks alot for the reply and amazing explanation!

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by Rich@VeritasPrep » Wed Jun 30, 2010 2:18 am
My pleasure :)
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by raunakrajan » Wed Jun 30, 2010 3:04 am
Hi Rich,I still dint get how did you get 51 pairs pls help me out with it!
Thanks
Raunak

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by Rich@VeritasPrep » Wed Jun 30, 2010 3:16 am
It's actually not 51 pairs. It's 51 numbers total in the list:

1+3+5+7+...+99+101

1 is the 1st positive odd number
3 is the 2nd positive odd number
5 is the 3rd positive odd number
...
2n-1 is the nth positive odd number
...
101 is the 51st positive odd number (since 2n-1 = 2(51) - 1 = 101)

Since there are 51 numbers in the list, there will be 25 pairs of numbers plus one left over.

Make sense?
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by raunakrajan » Wed Jun 30, 2010 3:48 am
yes thanks a ton Rich! :)