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19 Mar 2017, 18:09
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
90% (01:04) correct
10%
(01:28)
wrong
based on 149
sessions
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How many possible 6-digit code numbers can be formed from three a, two b, and one c?
A. 40
B. 50
C. 60
D. 70
E. 80
A. 40
B. 50
C. 60
D. 70
E. 80
19 Mar 2017, 20:54
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MathRevolution
----------ASIDE-----------------------
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:
If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]
So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
------NOW ONTO THE QUESTION----------
We want to arrange AAABBC
There are 6 letters in total
There are 3 identical A's
There are 2 identical B's
So, the total number of possible arrangements = 6!/[(3!)(2!)]
= 60
RELATED VIDEO
General Discussion
19 Mar 2017, 20:46
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Total Number of 6-digit code numbers = 6 ! / (3! * 2!) = 60
Answer is C. 60
Answer is C. 60
