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18 Mar 2024, 05:39
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
47% (03:10) correct
53%
(02:00)
wrong
based on 81
sessions
History
Date
Time
Result
Not Attempted Yet
How many 4-letter words can be created by using the letters of the word “ACHIEVED”?
A. 180
B. 510
C. 840
D. 1020
E. 1680
A. 180
B. 510
C. 840
D. 1020
E. 1680
Hi, could someone help me with this question found here
https://www.youtube.com/watch?v=HZH6V_JmhaI at 26:41.
Why does the author do 6C2 instead of 6*5? As can be seen he did that successfully for scenario 1 - and not sure why it is not applicable to scenario 2.
Thank you very much in advance.
Screenshot 2024-03-18 203757.png [ 254.83 KiB | Viewed 1814 times ]
https://www.youtube.com/watch?v=HZH6V_JmhaI at 26:41.
Why does the author do 6C2 instead of 6*5? As can be seen he did that successfully for scenario 1 - and not sure why it is not applicable to scenario 2.
Thank you very much in advance.
Attachment:
Screenshot 2024-03-18 203757.png [ 254.83 KiB | Viewed 1814 times ]
18 Mar 2024, 11:48
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unicornilove
We can divide the solution into two cases. Each case involves two steps -
a) Selecting the alphabet
b) Arranging selected alphabets
Case 1: When no alphabets repeat
a) Selecting the alphabet
The word "ACHIEVED" contains of 7 unique alphabets. From these 7 alphabets, we have to select 4 alphabets.
The selection can be done in \(^7C_4\) ways
b) Arranging selected alphabets
Each of the selected alphabets is unique. Hence the selected alphabets can be arranged in 4! ways.
Sub Total : \(^7C_4 * 4! = 840\) ways
Case 2: When only one alphabet repeats
a) Selecting the alphabet
The word "ACHIEVED" contains 7 unique alphabets. The only alphabet that repeats is E.
After the pair of repeating 'E's are selected, we need to select 2 more alphabets from the remaining 6 unique alphabets ('A', 'C', 'H', 'I', 'V', and 'D')
The selection can be done in \(^6C_2\) ways
b) Arranging selected alphabets
In the selected result, two alphabets repeat and two are unique. Hence the number of ways the four alphabets can be arranged is \(\frac{4! }{2!}\) ways
Sub Total : \(^6C_2* \frac{4!}{2!} = 180\) ways
The total number of ways is the sum of the number of ways obtained in Case 1 and the number of ways obtained in Case 2
Total = 840 + 180 = 1020
Option D
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Screenshot 2024-03-19 001751.png [ 174.42 KiB | Viewed 1997 times ]
02 Dec 2024, 00:56
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I have solved by creating two cases:
Case 1: all distinct alphabets = 840
Case 2:two E's and two distinct alphabets = 180
Total number of cases = 1020
Smart Prep (Tutor)
Case 1: all distinct alphabets = 840
Case 2:two E's and two distinct alphabets = 180
Total number of cases = 1020
Smart Prep (Tutor)
einstein801
