AAPL wrote:Manhattan GMAT

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

OA D.

There are 3 cases to consider:
case 1) numbers of the form XXY (1st and 2nd digits are the same and the 3rd digit is different)
case 2) numbers of the form XYX (1st and 3rd digits are the same and the 2nd digit is different)
case 3) numbers of the form YXX (2nd and 3rd digits are the same and the 1st digit is different)

Case 1: XXY
The 1st digit can be selected in 9 ways (1,2,3,4,5,6,7,8 or 9)
The 2nd digit can be selected in 1 way (since it must MATCH the 1st digit selected)
The 3rd digit can be selected in 8 ways (it can be any of the 9 digits EXCEPT the digit selected as the 1st and 2nd digits)
By the Fundamental Counting Principle (FCP), we can complete the 3 stages in (9)(1)(8) ways (= 72 ways)

Case 2: XYX
The 1st digit can be selected in 9 ways (1,2,3,4,5,6,7,8 or 9)
The 2nd digit can be selected in 8 ways (it can be any of the 9 digits EXCEPT the digit selected as the 1st digit)
The 3rd digit can be selected in 1 way (since it must MATCH the 1st digit selected)
By the Fundamental Counting Principle (FCP), we can complete the 3 stages in (9)(8)(1) ways (= 72 ways)

Case 3: YXX
The 1st digit can be selected in 9 ways (1,2,3,4,5,6,7,8 or 9)
The 2nd digit can be selected in 8 ways (it can be any of the 9 digits EXCEPT the digit selected as the 1st digit)
The 3rd digit can be selected in 1 way (since it must MATCH the 2nd digit selected)
By the Fundamental Counting Principle (FCP), we can complete the 3 stages in (9)(8)(1) ways (= 72 ways)

TOTAL number of outcomes = 72 + 72 + 72 = 216

Answer: D



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Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should ... 67256.html
- https://www.beatthegmat.com/counting-pr ... 44302.html
- https://www.beatthegmat.com/picking-a-5 ... 73110.html
- https://www.beatthegmat.com/permutation ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- https://www.beatthegmat.com/combinatori ... 73194.html
- https://www.beatthegmat.com/arabian-hor ... 50703.html
- https://www.beatthegmat.com/sub-sets-pr ... 73337.html
- https://www.beatthegmat.com/combinatori ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-se ... 71047.html
- https://www.beatthegmat.com/combinatori ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p ... 71001.html
- https://www.beatthegmat.com/permutation ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladi ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent

AAPL wrote:Manhattan GMAT

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

We have 9 digits (1 to 9, since no digit can be 0) to form such a 3-digit number. Since two of the digits are the same and the third is not, we are using 2 of the 9 digits, and there are 9C2 = (9x8)/2 = 36 ways to choose 2 digits.

Now, let's say the two digits we choose are 1 and 2, we can have 112, 121, 211, 221, 212, and 122. So for each set of 2 digits we choose, there are 6 such 3-digit numbers. Since there are 36 sets, there are 36 x 6 = 216 such numbers.

Answer: D