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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)?
$$ \\ $$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))
$$\mathrm{The}\:\mathrm{required}\:\mathrm{probability}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\left(\:\mathrm{the}\:\mathrm{number}\right. \\ $$$$\mathrm{of}\:\mathrm{permutations}\:\mathrm{with}\:\mathrm{2}\:\mathrm{a}'\mathrm{s}\:\mathrm{together}\:\mathrm{and}\:\mathrm{2}\:\mathrm{a}'\mathrm{s} \\ $$$$\left.\mathrm{separated}\right)\mathrm{divided}\:\mathrm{by}\:\mathrm{the}\:\mathrm{total}\:\mathrm{numbers}\:\mathrm{of} \\ $$$$\mathrm{permutations} \\ $$$$\left(\bullet\right)\:\mathrm{The}\:\mathrm{permutations}\:\mathrm{with}\:\mathrm{2}\:\mathrm{a}'\mathrm{s}\:\mathrm{together}\: \\ $$$$\mathrm{and}\:\mathrm{2}\:\mathrm{a}'\mathrm{s}\:\mathrm{separated}\:\mathrm{can}\:\mathrm{end}\:\mathrm{with}\:\mathrm{b}\:\mathrm{or}\:\mathrm{ba}\:\mathrm{or} \\ $$$$\mathrm{baa}\:. \\ $$$$\left(\mathrm{1}\right)\mathrm{The}\:\mathrm{permutations}\:\mathrm{which}\:\mathrm{end}\:\mathrm{with}\:\mathrm{b}\:\mathrm{can}\: \\ $$$$\mathrm{be}\:\mathrm{formed}\:\mathrm{by}\:\mathrm{combining}\:\mathrm{the}\:\mathrm{character}\:\mathrm{or} \\ $$$$\mathrm{string}\::\:\mathrm{aab},\:\mathrm{ab},\:\mathrm{ab}\:\mathrm{and}\:\mathrm{5b}'\mathrm{s}\:.\:\mathrm{The}\:\mathrm{number} \\ $$$$\mathrm{of}\:\mathrm{these}\:\mathrm{is}\:\mathrm{C}_{\mathrm{1}} ^{\:\mathrm{8}} \:×\:\mathrm{C}_{\mathrm{2}} ^{\:\mathrm{7}} \:=\:\mathrm{168} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{The}\:\mathrm{permutations}\:\mathrm{which}\:\mathrm{end}\:\mathrm{in}\:\mathrm{baa}\:\mathrm{be} \\ $$$$\mathrm{formed}\:\mathrm{by}\:\mathrm{combining}\:\mathrm{the}\:\mathrm{characters}\:\mathrm{or} \\ $$$$\mathrm{string}\::\:\mathrm{aab},\:\mathrm{ab}\:\mathrm{6b}'\mathrm{s}\:\mathrm{and}\:\mathrm{the}\:\mathrm{final}\:\mathrm{a}\:. \\ $$$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{these}\:\mathrm{is}\:\mathrm{C}_{\:\mathrm{1}} ^{\:\mathrm{8}} \:×\:\mathrm{C}_{\mathrm{1}} ^{\:\mathrm{7}} \:=\:\mathrm{56} \\ $$$$\:\left(\mathrm{3}\right)\:\mathrm{The}\:\mathrm{permutations}\:\mathrm{which}\:\mathrm{end}\:\mathrm{in}\:\mathrm{baa}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{formed}\:\mathrm{by}\:\mathrm{combining}\:\mathrm{the}\:\mathrm{characters}\:\mathrm{of}\:\mathrm{string} \\ $$$$\mathrm{ab},\:\mathrm{ab},\:\mathrm{6b}'\mathrm{s}\:\mathrm{and}\:\mathrm{the}\:\mathrm{final}\:\mathrm{aa}.\: \\ $$$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{these}\:\mathrm{is}\:\mathrm{C}_{\mathrm{2}} ^{\:\mathrm{8}} \:=\:\mathrm{28} \\ $$$$\left(\mathrm{4}\right)\mathrm{The}\:\mathrm{total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{permutations}\:\mathrm{is} \\ $$$$\:\mathrm{C}_{\mathrm{4}} ^{\:\mathrm{12}} \:=\:\mathrm{495} \\ $$$$ \\ $$$$\mathrm{So}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{is}\:\frac{\mathrm{168}+\mathrm{56}+\mathrm{28}}{\mathrm{495}}\:=\:\frac{\mathrm{28}}{\mathrm{55}} \\ $$
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))
$$\Box{aa}\boxdot{a}\boxdot{a}\Box \\ $$$$\boxdot={one}\:{or}\:{more}\:{b}'{s} \\ $$$$\Box={zero}\:{or}\:{more}\:{b}'{s} \\ $$$$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…\right)^{\mathrm{2}} \left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…\right)^{\mathrm{2}} ={x}^{\mathrm{2}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{3}} ^{{k}+\mathrm{3}} {x}^{{k}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{8}} \:{term}\:{is}\:{at}\:{k}=\mathrm{6}:\:{C}_{\mathrm{3}} ^{\mathrm{6}+\mathrm{3}} \\ $$$${number}\:{of}\:{valid}\:{words}:\:{C}_{\mathrm{3}} ^{\mathrm{6}+\mathrm{3}} ×\frac{\mathrm{3}!}{\mathrm{2}!}=\mathrm{252} \\ $$$${total}\:{number}\:{of}\:{words}:\:\frac{\mathrm{12}!}{\mathrm{4}!\mathrm{8}!}=\mathrm{495} \\ $$$${p}=\frac{{C}_{\mathrm{3}} ^{\mathrm{6}+\mathrm{3}} ×\frac{\mathrm{3}!}{\mathrm{2}!}}{\frac{\mathrm{12}!}{\mathrm{4}!\mathrm{8}!}}=\frac{\mathrm{252}}{\mathrm{495}}=\frac{\mathrm{28}}{\mathrm{55}} \\ $$

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