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Faster way to solve this problem

by mchaubey » Sat May 01, 2010 12:03 pm
Of the 3 digit intergers greater than 700, how many have 2 digits that are equal to each other and the remaining digit different from the other 2?

1. 90
2. 82
3. 80
4. 45
5.. 36

Got the answer but took a long time any faster way to do the same.

Thanks,
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by tpr-becky » Sat May 01, 2010 1:15 pm
This is a permutation problem so we just have to add up what is possible.

You have two scenarios - either the first number is part of the pair or it isn't.

When the first number is part of the pair you have 3 available for the first digit (7,8&9), then only one number will match those and there are 9 numbers available for the third digit. so 3(1)(9)=27. you have to multiply this by two becuase switching the 2nd and third digits will give you different numbers so we are up to 54.

The second scenario is when the first digit is not part of the pair that are equal to each other. We still have three digits available for the hundreds digit then 9 for the second digit and only 1 will match - for (3)(9)(1)=27 We dont' multiply these by 2 becuase that would create the same number (711/711). for a total of 81.

But, in the second scenario we included 700 and the question clearly says the integers greater than 700 so we have to 1 from the total to get 80. Answer is C[spoiler][/spoiler]
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by clock60 » Sat May 01, 2010 1:22 pm
mchaubey wrote:Of the 3 digit intergers greater than 700, how many have 2 digits that are equal to each other and the remaining digit different from the other 2?

1. 90
2. 82
3. 80
4. 45
5.. 36

Got the answer but took a long time any faster way to do the same.

Thanks,
i got C-80, my thinking
first with numbers starting with 7
7xx where x is any digit but 0 and 7 (as >700, and 777 is not possible), for x we have 10-2=8 options
7x7 where x is any digit but 7 and 10-1=9 options
77x again 9 options
and together
8+9+9=26 numbers

8xx-x any number but 8 as 888 is not possible so 9 options
8x8- 9 options
88x-9 options
together
9+9+9=27

9xx all but 9
9x9 all but 9
99x all but 9
together
9+9+9=27

totaling=26+27+27=80
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by susantaiitk » Mon May 03, 2010 8:29 am
since there are less numbers it can be done without formula

7 XX = 9 - 1 (as 00 can't be since we are asked to find numbers more than 700)
7X7 = 9
77X = 9
-----
27 - 1

similarly for 800 series and 900 series total are 27 and 27

so total = 27*3 - 1 = 80
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by kstv » Wed May 05, 2010 6:58 am
Of the 3 digit intergers greater than 700, how many have 2 digits that are equal to each other and the remaining digit different from the other 2?

Total no of three digits = 299 i.e 999-700
No of three digits where all the digits are diff = 3*9*8 = 216
hundreds place can have 7,8 or 9 so 3 ways
tens place = 9 and units place = 8
No of three digits where all digits are same = 3
i.e 777, 888 and 999
No of three digits where 2 digits are similar and one diff = 299-216-3 = 80
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