Combination of numbers

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Combination of numbers

by chayanika » Tue Sep 09, 2008 11:33 am
A state issues license plate with the following pattern : each plates has three digits (1 through 9). However , each digit must be equal to or greater than the one preceding. (For ex 121 is not permissible combination because the last 1 is less than the digit before it) How many different license plates can the state issue?
1) 210
2) 220
3) 222
4) 242
5) 165


Correct answer is 165

Please explain how to tackle this question
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Re: Combination of numbers

by parallel_chase » Tue Sep 09, 2008 1:13 pm
chayanika wrote:A state issues license plate with the following pattern : each plates has three digits (1 through 9). However , each digit must be equal to or greater than the one preceding. (For ex 121 is not permissible combination because the last 1 is less than the digit before it) How many different license plates can the state issue?
1) 210
2) 220
3) 222
4) 242
5) 165


Correct answer is 165

Please explain how to tackle this question
This question involves use of permutations as well as identifying the pattern.

options 1-9

Calculate permutations for only 100's

1*1*9 = 9 ways [place 1 at first place, 1 again second place, we have 9 options equal and greater than 1]

1*1*8 = 8 ways [place 1 at first place, 2 at second place, we have 8 options equal and greater than 2]

With similar reasoning below permutations follow with second place changing in increasing order from 3-9
1*1*7 = 7 ways
1*1*6 = 6 ways
1*1*5 = 5 ways
1*1*4 = 4 ways
1*1*3 = 3 ways
1*1*2 = 2 ways
1*1*1 = 1 ways

9+8+7+6+5+4+3+2+1 = 45 ways

For 200's follow the same as above

1*1*8 = 8 ways [ 2 at first place, 2 again at second place, we have 8 options for third place that are equal and greater than 2]

For 200's we have 8+7+6+5+4+3+2+1 = 36 ways

Now its time to identify the pattern

45 ways
45-9 = 36 ways
36-8 = 28 ways
28-7 = 21 ways
21-6 = 15 ways
15-5 = 10 ways
10-4 = 6 ways
6-3 = 3 ways
3-2 = 1 way

Therefore total possible number of plates are

45+36+28+21+15+10+6+3+1 = 165

This took me more than 5 minutes and I am not able to find a faster way to do this. Whats the source ???

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Re: Combination of numbers

by Stuart@KaplanGMAT » Tue Sep 09, 2008 1:26 pm
parallel_chase wrote:
chayanika wrote:A state issues license plate with the following pattern : each plates has three digits (1 through 9). However , each digit must be equal to or greater than the one preceding. (For ex 121 is not permissible combination because the last 1 is less than the digit before it) How many different license plates can the state issue?
1) 210
2) 220
3) 222
4) 242
5) 165


Correct answer is 165

Please explain how to tackle this question
This question involves use of permutations as well as identifying the pattern.

options 1-9

Calculate permutations for only 100's

1*1*9 = 9 ways [place 1 at first place, 1 again second place, we have 9 options equal and greater than 1]

1*1*8 = 8 ways [place 1 at first place, 2 at second place, we have 8 options equal and greater than 2]

With similar reasoning below permutations follow with second place changing in increasing order from 3-9
1*1*7 = 7 ways
1*1*6 = 6 ways
1*1*5 = 5 ways
1*1*4 = 4 ways
1*1*3 = 3 ways
1*1*2 = 2 ways
1*1*1 = 1 ways

9+8+7+6+5+4+3+2+1 = 45 ways

For 200's follow the same as above

1*1*8 = 8 ways [ 2 at first place, 2 again at second place, we have 8 options for third place that are equal and greater than 2]

For 200's we have 8+7+6+5+4+3+2+1 = 36 ways

Now its time to identify the pattern

45 ways
45-9 = 36 ways
36-8 = 28 ways
28-7 = 21 ways
21-6 = 15 ways
15-5 = 10 ways
10-4 = 6 ways
6-3 = 3 ways
3-2 = 1 way

Therefore total possible number of plates are

45+36+28+21+15+10+6+3+1 = 165

This took me more than 5 minutes and I am not able to find a faster way to do this. Whats the source ???
I'm also skeptical about the source, since it seems time consuming.

However, you could have saved a LOT of time if you had started at the top rather than the bottom:

if the first digit is 9, there's only 1 plate: 999
if the first digit is 8, there are 3 plates: 888, 889, 899
if the first digit is 7, there are 6 plates: 777, 778, 779, 788, 789, 799
if the first digit is 6, there are 10 plates: 666, 667, 668, 669, 677, 678, 679, 688, 689, 699

Now if we see the pattern: 1, 3, 6, 10.. increasing by 1 more each time, we project:

1, 3, 6, 10, 15, 21, 28, 36, 45 and them add them up, we get 165 total.

If you saw this approach right away, you could solve in under 2 minutes. I'm sure that there's an "elegant" approach, but I'm not sure that it's any quicker (especially when you factor in the time it takes to find the approach).
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Re: Combination of numbers

by Stockmoose16 » Tue Sep 09, 2008 2:50 pm
Stuart Kovinsky wrote:
parallel_chase wrote:
chayanika wrote:A state issues license plate with the following pattern : each plates has three digits (1 through 9). However , each digit must be equal to or greater than the one preceding. (For ex 121 is not permissible combination because the last 1 is less than the digit before it) How many different license plates can the state issue?
1) 210
2) 220
3) 222
4) 242
5) 165


Correct answer is 165

Please explain how to tackle this question
This question involves use of permutations as well as identifying the pattern.

options 1-9

Calculate permutations for only 100's

1*1*9 = 9 ways [place 1 at first place, 1 again second place, we have 9 options equal and greater than 1]

1*1*8 = 8 ways [place 1 at first place, 2 at second place, we have 8 options equal and greater than 2]

With similar reasoning below permutations follow with second place changing in increasing order from 3-9
1*1*7 = 7 ways
1*1*6 = 6 ways
1*1*5 = 5 ways
1*1*4 = 4 ways
1*1*3 = 3 ways
1*1*2 = 2 ways
1*1*1 = 1 ways

9+8+7+6+5+4+3+2+1 = 45 ways

For 200's follow the same as above

1*1*8 = 8 ways [ 2 at first place, 2 again at second place, we have 8 options for third place that are equal and greater than 2]

For 200's we have 8+7+6+5+4+3+2+1 = 36 ways

Now its time to identify the pattern

45 ways
45-9 = 36 ways
36-8 = 28 ways
28-7 = 21 ways
21-6 = 15 ways
15-5 = 10 ways
10-4 = 6 ways
6-3 = 3 ways
3-2 = 1 way

Therefore total possible number of plates are

45+36+28+21+15+10+6+3+1 = 165

This took me more than 5 minutes and I am not able to find a faster way to do this. Whats the source ???
I'm also skeptical about the source, since it seems time consuming.

However, you could have saved a LOT of time if you had started at the top rather than the bottom:

if the first digit is 9, there's only 1 plate: 999
if the first digit is 8, there are 3 plates: 888, 889, 899
if the first digit is 7, there are 6 plates: 777, 778, 779, 788, 789, 799
if the first digit is 6, there are 10 plates: 666, 667, 668, 669, 677, 678, 679, 688, 689, 699

Now if we see the pattern: 1, 3, 6, 10.. increasing by 1 more each time, we project:

1, 3, 6, 10, 15, 21, 28, 36, 45 and them add them up, we get 165 total.

If you saw this approach right away, you could solve in under 2 minutes. I'm sure that there's an "elegant" approach, but I'm not sure that it's any quicker (especially when you factor in the time it takes to find the approach).
Would a question like this really be on the GMAT? It's much too hard for most people to even begin to understand. I would've just multiplied 10*9*8 (the number of digits remaining, after each pick).

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by Ian Stewart » Wed Sep 10, 2008 6:25 am
The question can be answered quite quickly if you look at it in the right way. If we choose three different digits, there is only one order we can use (increasing order), and we can choose three different digits in 9C3 ways. If two of the digits are the same, and one different, we can choose the two digits in 9C2 ways, and two orders work: either the smaller number is repeated, or the larger is. If all the digits are identical, we can have 9C1 different numbers. So the answer will be 9C3 + 2*9C2 + 9C1 = 165.
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by pseudononymous » Wed Sep 10, 2008 7:32 am
9*1 + 8*2 + 7*3 + 6*4 + 5*5 + 4*6 + 3*7 + 2*8 = 1*9 = 165

How many 2-digit #s end in 9? [19,29,39,49,59,69,79,89,99 = 9]. How many 2-digit #s end in 8? [18,28,38,48,58,68,78,88 = 8], etc.

If the 2nd digit is 9, there's only 1 possible values for the last digit [9]. If the 2nd digit is 8, there's only 2 possible values for the last digit [8,9], etc.

Hence, 9*1 + 8*2 + etc.