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

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

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

error   error  I saw an answer for this, but it has been deleted....maybe because it was incorrect,,,   sorry

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.

Hi Rosala,

I have anwered on the original thread :)

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