set 25 q 11

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radhika1306: Master | Next Rank: 500 Posts; Posts: 144; Joined: Fri Apr 13, 2007 2:25 am

by radhika1306 » Mon Sep 17, 2007 9:58 am

Of the three-digit positive integers that have no digits equal
to zero, how many have two digits that are equal to each
other and the remaining digit different from the other two?
A. 24
B. 36
C. 72
D. 144
E. 216

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Source: Beat The GMAT — Problem Solving | Mon Sep 17, 2007 9:58 am

kajcha: Master | Next Rank: 500 Posts; Posts: 321; Joined: Tue Aug 28, 2007 5:42 am; Thanked: 1 times

by kajcha » Mon Sep 17, 2007 10:08 am

Lets assume no can be written as xxy

for first position you can have 9 different nos
Second position you would have only 1 possibility
third position you can have 8 possibility

so total is 9*8 = 72

Positions of x and y are interchangeable so no can be xxy, xyx, yxx - 3 ways

Total = 72*3 = 216

What's the OA

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radhika1306: Master | Next Rank: 500 Posts; Posts: 144; Joined: Fri Apr 13, 2007 2:25 am

by radhika1306 » Mon Sep 17, 2007 10:10 am

thanks a ton

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samirpandeyit62: Master | Next Rank: 500 Posts; Posts: 460; Joined: Sun Mar 25, 2007 7:42 am; Thanked: 27 times

by samirpandeyit62 » Mon Sep 17, 2007 10:27 am

I think ans should be E

say the repeated digit br R & the other digit be N

there are 3 possible arrangements for the nos

RRN
RNR
NRR

now if we tak case 1 we have

1st repeated digit R can be selected in 9 ways (0 excluded)
2nd repeated digit can be selected in 1 ways (same as we selected for first)
so 3rd digit i,e N can be selected in 8 ways

so total nos of ways is 9*1*8 =72 ways

same will come out for the other two arrangements

so total is 72+72+72 =216 ways

Regards
Samir

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radhika1306: Master | Next Rank: 500 Posts; Posts: 144; Joined: Fri Apr 13, 2007 2:25 am

by radhika1306 » Tue Sep 18, 2007 7:45 am

OA is E

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