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Sid intended to type a seven-digit number, but the two 3's h
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19 Nov 2019, 11:15
19 Nov 2019, 11:15
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Question Stats:
76% (01:23) correct
24% (01:28) wrong based on 25 sessions
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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
A. 10
B. 16
C. 21
D. 24
E. 27
Re: Sid intended to type a seven-digit number, but the two 3's h
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22 Nov 2019, 11:46
22 Nov 2019, 11:46
Expert Reply
Simple 7C2 = 21 ways.
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Joined: 10 Apr 2015
Posts: 6218
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Sid intended to type a seven-digit number, but the two 3's h
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20 Jul 2020, 06:40
20 Jul 2020, 06:40
1
huda wrote:
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
A. 10
B. 16
C. 21
D. 24
E. 27
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
Aside: The video below explains how to quickly calculate combinations (like 6C2) in your head
TOTAL number of possible outcomes = 6 + 15 = 21
Answer: C
Cheers,
Brent
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