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Combinatorics problem

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by sohrabkalra » Wed Oct 12, 2011 1:42 am
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
C)36
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by shankar.ashwin » Wed Oct 12, 2011 1:54 am
N(Exactly 2) = N(total) - N(All different) - N(all same)

Total = 700 - 999 = 299
All different = 3*9*8 = 216
All same = 777,888,999

N(Exactly 2) = 299-216-3 = 80
sohrabkalra wrote: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
C)36
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by Proleefeek111 » Wed Oct 12, 2011 2:27 am
sohrabkalra wrote: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
C)36
[/quote]

711 - 799 : 9
717 - 797 : 9 - 1= 8 (xcluding 777)
771 - 779 : 9 - 1 = 8 (xcluding 777)

Total = 25

the above pattern will repeat in the sample space (801-899) and (901 -999). hence total = 25*3 =75

770,880,990 - 3
800, 900 -2
Thus total combinations = 75 + 3 + 2 = 80.
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by sohrabkalra » Wed Oct 12, 2011 2:33 am
the answer is 80 !

I was following ashiwin's approach , but was doing a small overglance in the no of digits leading to a wrong calculation of ALL DIFFERENT !

Thanks
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