Question

$\red{\underline{\sf{Explain\:\:properly.}}}$
In how many ways the 26 letters of the English alphabet can permuted so that none of the word SUN , TEA , BOY or GIRL occurs?

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Answer 1

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!

Answer 2

Explanation:

Ways the 26 alphabets can be permuted so that SUN , TEA , BOY or GIRL occurs = 26! - [3(24!) + 23!] + [3(22!) + 3(21!)] - [20! + 3(19!)] - 17!

Step-by-step explanation:

Given: In how many ways the 26 letters of the English alphabet can be permuted so that none of the word SUN , TEA , BOY or GIRL occurs.

Solution:

Total number of permutations possible with 26 alphabets, N = 26!

Total number of permutations possible with 26 alphabets where SUN occurs always,

N(A) = (26 - 3 + 1)! = 24!

Similarly, for TEA, N(B) = 24!

for BOY, N(C) = 24!

for GIRL, N(D) = 23!

Total number of permutations possible with 26 alphabets where SUN and TEA occurs always, N(AB) = (26 - 3 - 3 +2)! = 22!

Similarly, for SUN and BOY, N(AC) = 22!,

for SUN and GIRL, N(AD) = 21!,

for TEA and BOY, N(BC) = 22!

for TEA and GIRL, N(BD) = 21!

for BOY and GIRL, N(CD) = 21!

Total number of permutations possible with 26 alphabets where SUN, TEA and BOY 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 SUN, TEA, BOY and GIRL occurs always, N(ABCD) = (26 - 3 - 3 - 3 - 4 + 4)! = 17!

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!

Hope it will help you sis☺️