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Digits

by beater » Fri Feb 13, 2009 7:57 pm
Of the 3-digit integers greater than 700, how many have 2 digits that are equal to each other and the remaining digit different from the other 2 ?

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

Could someone please show me how to solve this using combinatorics
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by DanaJ » Fri Feb 13, 2009 10:27 pm
https://www.beatthegmat.com/counting-int ... 16933.html
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Re: Digits

by billzhao » Sat Feb 14, 2009 1:31 am
beater wrote:Of the 3-digit integers greater than 700, how many have 2 digits that are equal to each other and the remaining digit different from the other 2 ?

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

Could someone please show me how to solve this using combinatorics
There are three situations:
1.All the digits are the same.
2.Two of the three are the same.
3.Every digit is distinct.

So we can use subtraction.

Total number of 3-digit integer: 3*10*10 = 300
Total number of 3-digit integer with all the digits are the same: 3*1*1=3 (i.e. 777,888,999)
Total number of 3-digit integer with every digit is distinct: 3*9*8=216
Total number of 3-digit integer with 2 digits that are equal to each other: 300-3-216=81

Then we need to exclude 700. So the answer is 81-1=80
Yiliang
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