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NUMBER PROPERTIES

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by amitdgr » Wed Sep 17, 2008 9:47 pm
Of the three digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?

A) 90
B) 82
C) 80
D) 45
E)36

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Re: NUMBER PROPERTIES

by justpal » Wed Sep 17, 2008 10:47 pm
I go with answer choice C = 80

calculation
707,711,717,722,727,733,737,744,747,755,757,766,767,788,787,797,799,770,771,772,773,774,775,776,778,779....Do not include 777 and 888 and 999 as question ask for only 2 digits to be same....

this will be true for 800 series and 900 series......

The count for all of them comes to 80.....

Does anyone knows the short cut....
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by calpol » Thu Sep 18, 2008 1:44 am
Using the counting principle:
7 - - -> This can be filled in 8 ways(Leaving 7 and 0, as 777 and 700 does not fit in the range)
7 - 7 -> This can be filled in 9 ways
7 7 - -> This can be filled in 9 ways.

So total number of digits that fit in the 700 series is 26, this count will be the same for 800 and 900 series and we need to add 800 and 900 too.
So 26+26+26+ 1 +1 = 80
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by aj5105 » Fri May 08, 2009 8:50 am
Can we use permutation here?

_ _ _ (three slots)

(I)

First slot: 7 or 8 or 9

Second slot: 7 or 8 or 9

Third slot: 0 to 6

OR

(II)

First slot: 7 or 8 or 9

Second slot: 0 to 6

Third slot: 7 or 8 or 9

(3 * 3 * 7 ) + (3 * 7 * 3) = 126

Where am I slipping?
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by vitaly » Fri May 08, 2009 9:22 am
>So 26+26+26+ 1 +1 = 80

How did u get +1+1?
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by m&m » Fri May 08, 2009 9:43 am
You can do this by knowing there are 299 possible number from 701 to 999 inclusive.

299- when all numbers are different - all numbers the same = ans

299 - 3C1*9C1*8C1 - 3 = 80


I think that's the easiest way
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