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Counting Problem

by anuptvm » Wed Dec 22, 2010 11:49 am
Hi,

I find the combinatorics problems intimidating. I know the basic formulae of Combination and Permutation, but its the application which stumps me at times. Anyways, here is a question that I came across. Could someone explain this.

How many odd three-digit integers greater than 800 are there such that all their digits are different?
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by Night reader » Wed Dec 22, 2010 12:04 pm
anuptvm wrote:Hi,

I find the combinatorics problems intimidating. I know the basic formulae of Combination and Permutation, but its the application which stumps me at times. Anyways, here is a question that I came across. Could someone explain this.

How many odd three-digit integers greater than 800 are there such that all their digits are different?
start from the last digit, as the constraint is odd integers:
2*9*5=90

2 (8 and 9 for hundredth digits)' 9 (0...9, excluding one digit of hundredth digit, or 9)' 5(only odd digits count, i.e. 1-3-5-7-9)

this is perms question rather combo
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by GMATGuruNY » Wed Dec 22, 2010 12:09 pm
anuptvm wrote:Hi,

I find the combinatorics problems intimidating. I know the basic formulae of Combination and Permutation, but its the application which stumps me at times. Anyways, here is a question that I came across. Could someone explain this.

How many odd three-digit integers greater than 800 are there such that all their digits are different?
From 800-899:
Number of choices for hundreds digit = 1 (must be 8, giving us 1 choice)
Number of choices for units digit = 5 (we could use 1, 3, 5, 7, or 9)
Number of choices for tens digit = 8 (we could use any digit but the two already used)
Multiplying, we get 1*5*8 = 40 integers.

From 900-999:
Number of choices for hundreds digit = 1 (must be 9, giving us 1 choice)
Number of choices for units digit = 4 (since we can't reuse 9, we could use 1, 3, 5, or 7)
Number of choices for tens digit = 8 (we could use any digit but the two already used)
Multiplying, we get 1*4*8 = 32 integers.

Total possible integers = 40+32 = 72.
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by fskilnik@GMATH » Wed Dec 22, 2010 12:10 pm
anuptvm wrote:How many odd three-digit integers greater than 800 are there such that all their digits are different?
Hi anuptvm!

Let ABC be "your number", of course A is 8 or 9 and C is 1,3,5,7 or 9 but the problem is with 9, that cannot occur simultaneously for A and C, correct?!

To deal with this "trouble" easily, just separate in 2 cases:

First Case: C = 9, then A must be 8 and B may be any of the 8 other digits (except 8 and 9) :: 8 possibilities, then.

Second Case: C = 1,3, 5 or 7 , then A may be 8 or 9 and B may be any of the other 8 digits (except C and A) :: 4*2*8 possibilities, then.

First & Second Case are mutually exclusive, therefore 8+4*2*8 = 8(1+8) = 72.

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by Night reader » Wed Dec 22, 2010 12:43 pm
GMATGuruNY wrote:
anuptvm wrote:Hi,

I find the combinatorics problems intimidating. I know the basic formulae of Combination and Permutation, but its the application which stumps me at times. Anyways, here is a question that I came across. Could someone explain this.

How many odd three-digit integers greater than 800 are there such that all their digits are different?
From 800-899:
Number of choices for hundreds digit = 1 (must be 8, giving us 1 choice)
Number of choices for units digit = 5 (we could use 1, 3, 5, 7, or 9)
Number of choices for tens digit = 8 (we could use any digit but the two already used)
Multiplying, we get 1*5*8 = 40 integers.

From 900-999:
Number of choices for hundreds digit = 1 (must be 9, giving us 1 choice)
Number of choices for units digit = 4 (since we can't reuse 9, we could use 1, 3, 5, or 7)
Number of choices for tens digit = 8 (we could use any digit but the two already used)
Multiplying, we get 1*4*8 = 32 integers.

Total possible integers = 40+32 = 72.
thanks Mitch, again I approached combos mechanically :(
we need two perm intervals.
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by anshumishra » Wed Dec 22, 2010 12:46 pm
anuptvm wrote:Hi,

I find the combinatorics problems intimidating. I know the basic formulae of Combination and Permutation, but its the application which stumps me at times. Anyways, here is a question that I came across. Could someone explain this.

How many odd three-digit integers greater than 800 are there such that all their digits are different?
Let's make it even simpler.

Total number of integers greater than 800 such that all their digits are different = 2*9*8 = 144 (2 - Hundered's digit - either 8 or 9, 9 -Ten's digit all 9 digits except whatever is on hundred's digit, 8 - unit digit based on the same logic, that it shouldn't be what hundred or ten's digit is).

Now there should be exactly same number of odd digit and even digit between 800 and 999.
So, the number of odd digit = 144/2 = 72.

Thanks
Last edited by anshumishra on Wed Dec 22, 2010 1:45 pm, edited 1 time in total.
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by anuptvm » Wed Dec 22, 2010 12:49 pm
Many thanks to Mitch and Fabio. Both of you explained it succinctly. I guess I will have to do a few more of these problems to get a hang of it.

Warm Regards,
Anup
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by GMATGuruNY » Wed Dec 22, 2010 1:13 pm
anshumishra wrote:
anuptvm wrote:Hi,

I find the combinatorics problems intimidating. I know the basic formulae of Combination and Permutation, but its the application which stumps me at times. Anyways, here is a question that I came across. Could someone explain this.

How many odd three-digit integers greater than 800 are there such that all their digits are different?
Let's make it even simpler.

Total number of integers greater than 800 such that all their digits are different = 2*9*8 = 144 (2 - Hundered's digit - either 8 or 9, 9 -Ten's digit all 9 digits except whatever is on hundred's digit, 8 - unit digit based on the same logic, that it shouldn't be what hundred or ten's digit is).

Now there should be exactly same number of odd digit and even digit between 800 and 899.
So, the number of odd digit = 144/2 = 72.

Thanks
Very nice approach. Just be careful. This approach works because we have 1 odd option for the hundreds digit (9) and 1 even option (8). This approach would not work if we were asked to count the number of odd integers from 700-999 in which no digit was repeated, because then we would have 2 odd options for the hundreds digit (7 and 9) but only 1 even option (8).
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by anshumishra » Wed Dec 22, 2010 1:45 pm
GMATGuruNY wrote:
anshumishra wrote:
anuptvm wrote:Hi,

I find the combinatorics problems intimidating. I know the basic formulae of Combination and Permutation, but its the application which stumps me at times. Anyways, here is a question that I came across. Could someone explain this.

How many odd three-digit integers greater than 800 are there such that all their digits are different?
Let's make it even simpler.

Total number of integers greater than 800 such that all their digits are different = 2*9*8 = 144 (2 - Hundered's digit - either 8 or 9, 9 -Ten's digit all 9 digits except whatever is on hundred's digit, 8 - unit digit based on the same logic, that it shouldn't be what hundred or ten's digit is).

Now there should be exactly same number of odd digit and even digit between 800 and 899.
So, the number of odd digit = 144/2 = 72.

Thanks
Very nice approach. Just be careful. This approach works because we have 1 odd option for the hundreds digit (9) and 1 even option (8). This approach would not work if we were asked to count the number of odd integers from 700-999 in which no digit was repeated, because then we would have 2 odd options for the hundreds digit (7 and 9) but only 1 even option (8).
Agreed ! This is what you have to be careful about when you use symmetry !
Thanks for pointing that out here.

In that case between 800-999 equal even/odd non-repeated digit numbers.
Between 700-799 even/odd numbers would be in ratio 5:4 (So, a total of 40:32).
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