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
GPA: 2.8
WE:Engineering (Computer Software)
Re: How many 3 digit numbers can we make such that two of the digits are
[#permalink]
13 Jun 2021, 02:54
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!