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arithmetic prob

by nitug » Sat Aug 30, 2008 12:08 pm
hi guys,

i am looking for the method solving the que below, though was able to get the ans through my own combinations but made it longer than 2 mins for me.



How many odd three-digit integers greater than 800 are there such that all their digits are different?
• 40
• 56
• 72
• 81
• 104

thanks,
nitug
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Re: arithmetic prob

by parallel_chase » Sat Aug 30, 2008 12:17 pm
nitug wrote:hi guys,

i am looking for the method solving the que below, though was able to get the ans through my own combinations but made it longer than 2 mins for me.



How many odd three-digit integers greater than 800 are there such that all their digits are different?
• 40
• 56
• 72
• 81
• 104

thanks,
nitug
Here is the process:

for first digit we have 2 options (8,9)
for second digit we have 9 options (1,2,3,4,5,6,7,8,9,0)=10-1(which is used in the first option) = 9
for third digit we have 5 options (1,3,5,7,9).

we have to place 1 at the third digit.
2*8*1 = 16
first place can be arranged in two ways (8,9)
third place can be arranged in one way (1)
second place can be arranged in 8 ways.

Similarly we can arrange all the the three remaining odd integers i.e 3,5,7

16*4 = 64

we have to place 9 at the third digit.
1*8*1 = 8
first place can be arranged in one way (8)
third place can be arranged in one way (9)
second place can be arranged in 8 ways.

64+8 = 72

Let me know if you have any doubts.
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by pepeprepa » Sat Aug 30, 2008 12:22 pm
How many odd three-digit integers greater than 800 are there such that all their digits are different?

The first digit can be: 8, 9
The second digit can be: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
The third can be: 1, 3, 5, 7, 9

Number of possibilities: 2*9*5=90
(9 because we already chose 8 or 9 before and we do not want to have it again)

Ok now we have to substract the one who have 1 3 5 7 9 in double,
There are 9 of this type:
911, 933, 955, 977 (not 999 because we did not count it before), 811, 833, 855, 877, 899
[ I forgot 909,919,929,939,949,959,969,979,989 --> 72 indeed]

90-9=81

I think it's 81
Last edited by pepeprepa on Sat Aug 30, 2008 12:27 pm, edited 2 times in total.
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by nitug » Sat Aug 30, 2008 12:23 pm
thanks parallel_chase!!

thats a good strategy.

also , lemme know how can i improve my permuations and combinations skills. Cudnt even think of doing that way.

calculated by mentally making a map between numbers but took a longer time.:(


ans is indeed 72. :)
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by pepeprepa » Sat Aug 30, 2008 12:33 pm
I prefer Chase's method as well, that's short and efficient.
How did you approach it nitug?
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by parallel_chase » Sat Aug 30, 2008 12:34 pm
pepeprepa wrote:How many odd three-digit integers greater than 800 are there such that all their digits are different?

The first digit can be: 8, 9
The second digit can be: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
The third can be: 1, 3, 5, 7, 9

Number of possibilities: 2*9*5=90
(9 because we already chose 8 or 9 before and we do not want to have it again)

Number of possibilities: 2*9*5=90
2*10*5 = 100
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by pepeprepa » Sat Aug 30, 2008 12:38 pm
parallel_chase wrote:
pepeprepa wrote:How many odd three-digit integers greater than 800 are there such that all their digits are different?

The first digit can be: 8, 9
The second digit can be: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
The third can be: 1, 3, 5, 7, 9

Number of possibilities: 2*9*5=90
(9 because we already chose 8 or 9 before and we do not want to have it again)

Number of possibilities: 2*9*5=90
2*10*5 = 100
I believe my error is not there, that's 9 not 10 I wanted to put, I forgot 9 other possibilities that I have to substract to the total possibilities I found.
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by parallel_chase » Sat Aug 30, 2008 12:44 pm
pepeprepa wrote: I believe my error is not there, that's 9 not 10 I wanted to put, I forgot 9 other possibilities that I have to substract to the total possibilities I found.
Oh my bad, i thought you were considering total possibilities.
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Re: arithmetic prob

by sudhir3127 » Sat Aug 30, 2008 12:45 pm
nitug wrote:hi guys,

i am looking for the method solving the que below, though was able to get the ans through my own combinations but made it longer than 2 mins for me.



How many odd three-digit integers greater than 800 are there such that all their digits are different?
• 40
• 56
• 72
• 81
• 104

thanks,
nitug
its 72

first digit can be 8 or 9 ... 2 ways
middle digit can be any one from (0-9) other than the used one ..hence 9 ways
last digit can be in 4 ways ( 1 3 5 7)

thus its 2*9*4 = 72

hope that helps..
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re: pepeprepa

by nitug » Sat Aug 30, 2008 12:46 pm
i did a funny calculations :D

okay, lemme tell u.

started with 801,803,805,807, 809( total 5 numbers) and since for every even number in the second place, it will be 5 and there are four options for that(0,2,4,6), total comes out to be 5*4= 20,

similarly, for odd number in the middle place, 4 * 5 = 20

since now there will be one repitition in the digits with odd number at middle, so 4 options for each and middle digit options are 1,3,5,7,9( 5 options).

Now for numbers starting with 9.

4( total numbers made with each option of middle digit) * 5 ( total options with middle number as even)

3( total numbers made with each option of middle digit) * 4 ( total options with middle number as odd)


hope i explained it well.....!!
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by lunarpower » Sat Aug 30, 2008 10:49 pm
a basic rule to follow here:
whenever you're making sequential decisions (such as choosing the digits in this problem), PICK THE MOST RESTRICTED CHOICES FIRST.
this means that you want to pick the hundreds and units digits BEFORE the tens digit, because there are no restrictions on the tens digit at all, other than that it can't be the same as one of the digits already chosen).

also, there are only two possibilities for the hundreds digit, 8 and 9, and each of these possibilities has different effects on the units digit. therefore, it's best to consider these 2 possibilities separately.

NUMBERS STARTING WITH 8
there are 5 possibilities for the units digit (1, 3, 5, 7, 9).
once the units digit has been chosen, there are 8 possibilities for the tens digit: all possible digits, except the 2 that have already been chosen.
that's 5 x 8, or 40.

NUMBERS STARTING WITH 9
there are 4 possibilities for the units digit (1, 3, 5, 7), because 9 is now unavailable.
once the units digit has been chosen, there are 8 possibilities for the tens digit: all possible digits, except the 2 that have already been chosen.
that's 4 x 8, or 32.

therefore, there are 40 + 32 = 72 different numbers.
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