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can anyone pls answer the first one!!!??

 

https://web2.0calc.com/questions/permutations-and-combinations-lt-3

 Feb 10, 2018
edited by rosala  Feb 10, 2018
 #1
avatar+8842 
+2

Hey rosala, your link doesn't need the  " /new "  at the end...it should just be

 

https://web2.0calc.com/questions/permutations-and-combinations-lt-3

 

smiley

 Feb 10, 2018
 #2
avatar+11885 
+1

thanks hectictar for noticing..i appreciate it!thank you! smiley

 Feb 10, 2018
 #3
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error   error  I saw an answer for this, but it has been deleted....maybe because it was incorrect,,,   sorry

 Feb 10, 2018
edited by Guest  Feb 10, 2018
edited by Guest  Feb 10, 2018
 #5
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No, this one wasn't.

Guest Feb 10, 2018
 #4
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1.Find the no. of permutations of letters a , b , c , d, e , f , g taken all at a time if neither  “beg” nor “cad” patterns appear in any word.

 

Permuatations of 7 letters =7! =5,040

The word "beg" appears in 7 - 3 + 1 = 5 positions

The remaining 4 letters permute in 4! =24 ways

5 x 24 = 120 pemutations

 

Similarly, the word"cad" appears in 7 -3 + 1=5 positions

The remaining 4 letters permute in 4! = 24 ways

5 x 24 = 120 permutations.

 

The combinations of the word "beg" and "cad" as one word "begcad" appears in 7 - 6 + 1 =2 positions only. Similarly, the word "cadbeg" also appears in 2 positions only for a total of 4 words. Each one of these words is excluded twice, once when it is included in the 120 permuations of "beg" and once in the permutations of the word "cad". So these 4 words should be added back to the total.

 

So, the total permuations left should be  =5,040 -120 - 120 + 4 =4,804 permutations.

Melody: You are pretty good at this kind of problem, so, please take a look at it.

 Feb 10, 2018
 #6
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What about this permunation?:

 

B-E-G-F-C-A-D

Guest Feb 11, 2018
 #8
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Certainly I can do that :))

NOTE: I have not looked at your answer. 

I have answered on Rosala's original thread as indicated below :)

Melody  Feb 11, 2018
 #7
avatar+106977 
0

Hi Rosala,

I have anwered on the original thread :)

 

https://web2.0calc.com/questions/permutations-and-combinations-lt-3#r9

 Feb 11, 2018

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