Sid intended to type a seven-digit number, but the two 3's he meant to type did not appear. What appeared instead was

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Sid intended to type a seven-digit number, but the two 3's he meant to type did not appear. What appeared instead was

by BTGmoderatorDC » Wed Jan 19, 2022 5:00 pm

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Sid intended to type a seven-digit number, but the two 3's he meant to type did not appear. What appeared instead was the five-digit number 52115. How many different seven-digit numbers could Sid have meant to type?

A. 10
B. 16
C. 21
D. 24
E. 27

OA C

Source: Magoosh

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Re: Sid intended to type a seven-digit number, but the two 3's he meant to type did not appear. What appeared instead wa

by [email protected] » Thu Jan 20, 2022 7:04 am
BTGmoderatorDC wrote:
Wed Jan 19, 2022 5:00 pm
Sid intended to type a seven-digit number, but the two 3's he meant to type did not appear. What appeared instead was the five-digit number 52115. How many different seven-digit numbers could Sid have meant to type?

A. 10
B. 16
C. 21
D. 24
E. 27

OA C

Source: Magoosh
There are two possible cases to examine here:
case i) the two 3's are adjacent
case ii) the two 3's are NOT adjacent

case i) the two 3's are adjacent
Let's add spaces to where the two 3's can appear: _5_2_1_1_5_
There are 6 possible spaces where the two adjacent 3's can appear
So there are 6 possible ways in which the two adjacent 3's can appear

case ii) the two 3's are NOT adjacent
Let's add spaces to where the two 3's can appear: _5_2_1_1_5_
There are 6 possible spaces where the two non-adjacent 3's can appear
We must choose 2 different spaces
Since the order in which we choose the two spaces does not matter, we can use combinations.
We can choose 2 of the 6 spaces in 6C2 ways
6C2 = (6)(5)/(2)(1) = 15
So there are 15 possible ways in which two non-adjacent 3's can appear
TOTAL number of possible outcomes = 6 + 15 = 21