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stne
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Find the number of "ARRANGEMENTS " that can be made taking 15 Sep 2012, 06:42
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1)Find the number of "ARRANGEMENTS " that can be made taking 4 letters from the word "ENTRANCE".

Please note it says Arrangements and Not selections so ENTR is Different from NTRE meaning order is important.

OA = 606

However I am not sure how to do it



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EvaJager
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Find the number of "ARRANGEMENTS " that can be made taking 15 Sep 2012, 07:42
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stne
1)Find the number of "ARRANGEMENTS " that can be made taking 4 letters from the word "ENTRANCE".

Please note it says Arrangements and Not selections so ENTR is Different from NTRE meaning order is important.

OA = 606

However I am not sure how to do it
Show more

As I understand, we have to make four-letter words (even meaningless) using the letters of the word ENTRANCE.
There are 8 letters. Since we have two E's and two N's, we should distinguish between the following cases:

1) Choose two E's and two N's (E,E,N,N) - 1 possibility - number of words - 4!/(2!*2!) = 6
2) Choose two E's and any two letters of the other remaining 5 distinct letters (just one N included) - 5C2 = 10 possibilities - number of words - 10*(4!/2!) = 120
3) Choose two N's and any two letters of the other remaining 5 distinct letters (just one E included) - 5C2 = 10 possibilities - number of words - 10*(4!/2!) = 120
(Identical reasoning for 2 and 3 above)
4) Choose any four letters from 6 distinct letters (just one E and one N included) - 6C2 = 15 possibilities - number of words - 15*4!= 360.

Total number of possible words 6 + 2*120 + 360 = 606.
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stne
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Find the number of "ARRANGEMENTS " that can be made taking 15 Sep 2012, 08:01
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EvaJager
stne
1)Find the number of "ARRANGEMENTS " that can be made taking 4 letters from the word "ENTRANCE".

Please note it says Arrangements and Not selections so ENTR is Different from NTRE meaning order is important.

OA = 606

However I am not sure how to do it

As I understand, we have to make four-letter words (even meaningless) using the letters of the word ENTRANCE.
There are 8 letters. Since we have two E's and two N's, we should distinguish between the following cases:

1) Choose two E's and two N's (E,E,N,N) - 1 possibility - number of words - 4!/(2!*2!) = 6
2) Choose two E's and any two letters of the other remaining 5 distinct letters (just one N included) - 5C2 = 10 possibilities - number of words - 10*(4!/2!) = 120
3) Choose two N's and any two letters of the other remaining 5 distinct letters (just one E included) - 5C2 = 10 possibilities - number of words - 10*(4!/2!) = 120
(Identical reasoning for 2 and 3 above)
4) Choose any four letters from 6 distinct letters (just one E and one N included) - 6C2 = 15 possibilities - number of words - 15*4!= 360.

Total number of possible words 6 + 2*120 + 360 = 606.
Show more

awesome got it , finally , Thank you



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!