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three-digit positive integer

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by mgmt_gmat » Fri Feb 12, 2010 2:51 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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by sanju09 » Fri Feb 12, 2010 3:56 am
mgmt_gmat wrote: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
There are 3! /2! = 3 different arrangements for a type a a b.

Let's talk restricted digits for a and b in a a b, we have 9 ways for each a and 8 for b, hence a a b in this form can be constructed in (9 × 9 × 8)/3 = 648/3 = [spoiler]216[/spoiler].

[spoiler]E[/spoiler]
The mind is everything. What you think you become. -Lord Buddha



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by abhi332 » Fri Feb 12, 2010 3:57 am
we have to select digits from 1 to 9

First Case:

Ist digit selected = 9C1 = 9
2nd digit seleted which is same as 1st one = 1C1 = 1
3rd digit selected from 8 numbers remaining = 8C1 = 8

Second Case:

Ist digit selected = 9C1 = 9
2nd digit seleted which is not same as 1st one = 8C1 = 8
3rd digit selected which is same as 2nd digit = 1C1 = 1


No of ways = 9*8 + 8*9 = 72+72 =144

Correct me if I am wrong
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by sanju09 » Fri Feb 12, 2010 4:05 am
abhi332 wrote:we have to select digits from 1 to 9

First Case:

Ist digit selected = 9C1 = 9
2nd digit seleted which is same as 1st one = 1C1 = 1
3rd digit selected from 8 numbers remaining = 8C1 = 8

Second Case:

Ist digit selected = 9C1 = 9
2nd digit seleted which is not same as 1st one = 8C1 = 8
3rd digit selected which is same as 2nd digit = 1C1 = 1


No of ways = 9*8 + 8*9 = 72+72 =144

Correct me if I am wrong
Or we can then again use the following method:

112 to 119 have 3 × 8 = 24 arrangements

221 to 229 have 3 × 8 = 24 arrangements (222 not taken)

332 to 339 have 3 × 8 = 24 arrangements (333 not taken)
.
.
.
.
.
991 to 998 have 3 × 8 = 24 arrangements

Hence, it'll be 24 × 9 = [spoiler]216[/spoiler].

[spoiler]E[/spoiler]
The mind is everything. What you think you become. -Lord Buddha



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by komal » Tue Feb 16, 2010 7:32 pm
mgmt_gmat wrote: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
1st digit can be chosen in 9 ways..
2nd digit can be chosen in 1 way
3rd digit can be chosen in 8 ways...

total ways = 9*1*8= 72 ways..and 3 chosen digits can be arranged among themselves such that 2 digits are same and 3rd is different in 3!/2 = 3 ways..

total poss 3 digit nos = 72*3=216 ways..
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by thephoenix » Tue Feb 16, 2010 8:45 pm
there are 9 digits avaible for selection

selection of 1 dig can be done in 9c1=9 ways
selection of 2nd one can be done in 1 ways(as its same as prev one)
selection of 3rd from the rest of the 8 in 8c1=8

tot=9*8=72

now no. of ways in which the digits can be arranged =3!/2!=3ways

tot=72*3=216
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