In how many ways can the letters of the english alphabets cans be arranged so that
Answers
Step-by-step explanation:
Total number of permutations possible with 26 alphabets, N = 26!
Total number of permutations possible with 26 alphabets where CAR occurs always, N(A) = (26 - 3 + 1)! = 24!
Similarly, for DOG, N(B) = 24!,
for PUN, N(C) = 24!,
for BYTE, N(D) = 23!
Total number of permutations possible with 26 alphabets where CAR and DOG occurs always, N(AB) = (26 - 3 - 3 +2)! = 22!
Similarly, for CAR and PUN, N(AC) = 22!,
for CAR and BYTE, N(AD) = 21!,
for DOG and PUN, N(BC) = 22!
for DOG and BYTE, N(BD) = 21!
for PUN and BYTE, N(CD) = 21!
Total number of permutations possible with 26 alphabets where CAR, DOG and PUN occurs always, N(ABC) = (26 - 3 - 3 - 3 + 3)! = 20!
N(ABD) = 19!, N(ACD) = 19!, and N(BCD) = 19!
Total number of permutations possible with 26 alphabets where CAR, DOG, PUN and BYTE occurs always, N(ABCD) = (26 - 3 - 3 - 3 - 4 + 4)! = 17!
Now we can apply the principle of Inclusion and Exclusion to calculate the number of possible pemutations where none of the given pattern occur,
The required number of permutations = N - [N(A) + N(B) + N(C) + N(D)] + [N(AB) + N(AC) + N(AD) + N(BC) + N(BD) + N(CD)] - [N(ABC) + N(ABD) + N(ACD) + N(BCD)] - N(ABCD)
= 26! - (24! + 24! +24! + 23!) + (22! + 22! +22! +21! + 21! + 21!) - (20! +19! +19! +19!) +17!
= 26! - [3(24!) + 23!] + [3(22!) + 3(21!)] - [20! + 3(19!)] + 17!