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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
How many 3 digit numbers can we make such that two of the digits are
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10 Jun 2021, 22:04
10 Jun 2021, 22:04
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How many 3 digit numbers can we make such that two of the digits are same and non of the digits equal zero?
A. 60
B. 72
C. 150
D. 216
E. 280
A. 60
B. 72
C. 150
D. 216
E. 280
Moderator
Joined: 02 Jan 2020
Status:GRE Quant Tutor
Posts: 1111
Given Kudos: 9
Location: India
Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
GPA: 2.8
WE:Engineering (Computer Software)
Re: How many 3 digit numbers can we make such that two of the digits are
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13 Jun 2021, 02:54
13 Jun 2021, 02:54
1
This is a selection problem
Let's select the two values of three which are going to be the same. This can be done in 3C2 ways
=> \(\frac{3!}{(2!*1!)}\) = 3 ways
Now the digit which is going to be the same can be selected in 9 ways
(we have digits from 1 to 9 to select as digits cannot be 0)
The digit which is different can be select in 8 ways. (all numbers from 1 to 9 except the one selected for the common digits)
So, in total we have 3 * 9 * 8 = 216 ways
So, answer will be D
Hope it helps!
Let's select the two values of three which are going to be the same. This can be done in 3C2 ways
=> \(\frac{3!}{(2!*1!)}\) = 3 ways
Now the digit which is going to be the same can be selected in 9 ways
(we have digits from 1 to 9 to select as digits cannot be 0)
The digit which is different can be select in 8 ways. (all numbers from 1 to 9 except the one selected for the common digits)
So, in total we have 3 * 9 * 8 = 216 ways
So, answer will be D
Hope it helps!
gmatclubot
Re: How many 3 digit numbers can we make such that two of the digits are [#permalink]
13 Jun 2021, 02:54
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