Consider-a-12-letter-word-made-up-of-8-b-s-and-4-a-s-what-is-the-probability-that-randomly-shuffling-its-letters-lets-exactly-two-a-s-come-together-and-two-other-a-s-be-separated-as-in-the-followin

Question Number 134498 by bemath last updated on 04/Mar/21

Consider a 12-letter word made up of 8 b’s and 4 a’s. what is the probability that randomly shuffling its letters lets exactly two a’s come together and two other a’s be separated (as in the following example: baababbabbbb)?

Answered by EDWIN88 last updated on 04/Mar/21

The required probability is equal to( the number of permutations with 2 a′s together and 2 a′s separated)divided by the total numbers of permutations (•) The permutations with 2 a′s together and 2 a′s separated can end with b or ba or baa . (1)The permutations which end with b can be formed by combining the character or string : aab, ab, ab and 5b′s . The number of these is C_1 ^( 8) × C_2 ^( 7) = 168 (2) The permutations which end in baa be formed by combining the characters or string : aab, ab 6b′s and the final a . The number of these is C_( 1) ^( 8) × C_1 ^( 7) = 56 (3) The permutations which end in baa can be formed by combining the characters of string ab, ab, 6b′s and the final aa. The number of these is C_2 ^( 8) = 28 (4)The total number of permutations is C_4 ^( 12) = 495 So the probability is ((168+56+28)/(495)) = ((28)/(55))

The required probability is equal to (the number

of permutations with 2 a^{'} s together and 2 a^{'} s

separated) divided by the total numbers of

permutations

(∙) The permutations with 2 a^{'} s together

and 2 a^{'} s separated can end with b or ba or

baa .

(1) The permutations which end with b can

be formed by combining the character or

string : aab, ab, ab and {5 b}^{'} s . The number

of these is C_{1}^{8} \times C_{2}^{7} = 168

(2) The permutations which end in baa be

formed by combining the characters or

string : aab, ab {6 b}^{'} s and the final a .

The number of these is C_{1}^{8} \times C_{1}^{7} = 56

(3) The permutations which end in baa can be

formed by combining the characters of string

ab, ab, {6 b}^{'} s and the final aa .

The number of these is C_{2}^{8} = 28

(4) The total number of permutations is

C_{4}^{12} = 495

So the probability is \frac{168 + 56 + 28}{495} = \frac{28}{55}

Answered by mr W last updated on 04/Mar/21

□aa⊡a⊡a□ ⊡=one or more b′s □=zero or more b′s (1+x+x^2 +...)^2 (x+x^2 +x^3 +...)^2 =x^2 Σ_(k=0) ^∞ C_3 ^(k+3) x^k coef. of x^8 term is at k=6: C_3 ^(6+3) number of valid words: C_3 ^(6+3) ×((3!)/(2!))=252 total number of words: ((12!)/(4!8!))=495 p=((C_3 ^(6+3) ×((3!)/(2!)))/((12!)/(4!8!)))=((252)/(495))=((28)/(55))

a a ⊡ a ⊡ a

⊡ = o n e o r m o r e b^{'} s

= z e r o o r m o r e b^{'} s

{(1 + x + x^{2} + \dots)}^{2} {(x + x^{2} + x^{3} + \dots)}^{2} = x^{2} \underset{k = 0}{\sum^{\infty}} C_{3}^{k + 3} x^{k}

c o e f . o f x^{8} t e r m i s a t k = 6 : C_{3}^{6 + 3}

n u m b e r o f v a l i d w o r d s : C_{3}^{6 + 3} \times \frac{3!}{2!} = 252

t o t a l n u m b e r o f w o r d s : \frac{12!}{4! 8!} = 495

p = \frac{C_{3}^{6 + 3} \times \frac{3!}{2!}}{\frac{12!}{4! 8!}} = \frac{252}{495} = \frac{28}{55}

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