#1**+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

hectictar
Feb 10, 2018

#3**0 **

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

ElectricPavlov
Feb 10, 2018

edited by
Guest
Feb 10, 2018

edited by Guest Feb 10, 2018

edited by Guest Feb 10, 2018

#4**0 **

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.**

Guest Feb 10, 2018

#7**0 **

Hi Rosala,

I have anwered on the original thread :)

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

Melody
Feb 11, 2018