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Find-the-probability-that-a-student-arranging-the-letters-of-the-word-MATHEMATICS-will-make-all-the-vowels-be-together-in-any-arrangement-he-or-she-does-

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Question Number 56025 by pieroo last updated on 08/Mar/19
Find the probability that a student arranging the letters of the word MATHEMATICS will make all the vowels be together in any arrangement he or she does.
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{a}\:\mathrm{student} \\ $$$$\mathrm{arranging}\:\mathrm{the}\:\mathrm{letters}\:\mathrm{of}\:\mathrm{the}\:\mathrm{word} \\ $$$$\mathrm{MATHEMATICS}\:\mathrm{will}\:\mathrm{make}\:\mathrm{all}\:\mathrm{the} \\ $$$$\mathrm{vowels}\:\mathrm{be}\:\mathrm{together}\:\mathrm{in}\:\mathrm{any}\:\mathrm{arrangement} \\ $$$$\mathrm{he}\:\mathrm{or}\:\mathrm{she}\:\mathrm{does}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 08/Mar/19
A=2 C=1 E=1 H=1 I=1 M=2 S=1 T=2 (AAEI)CHMMSTT (((8!)/(2!×2!)))×((4!)/(2!))→it is the answer ((8×7×6×5×3×2)/1)×((4×3)/1)=20160×6=120960 discussion assumed vowel AAEI are Fabiquicked permutation for AAEI is=(4/(2!)) denominator 2! for two A permutation for (AAEI)CHMMSTT ((8!)/(2!×2!))→denominator 2!for two(02)M anither2! for two T so answer=((8!)/(2!2!))×((4!)/(2!)) yes i forgot to find probablity...MrW sir done it pls check...
$${A}=\mathrm{2}\:\:{C}=\mathrm{1}\:{E}=\mathrm{1}\:\:{H}=\mathrm{1}\:\:{I}=\mathrm{1}\:{M}=\mathrm{2}\:{S}=\mathrm{1}\:{T}=\mathrm{2} \\ $$$$\left({AAEI}\right){CHMMSTT} \\ $$$$\left(\frac{\mathrm{8}!}{\mathrm{2}!×\mathrm{2}!}\right)×\frac{\mathrm{4}!}{\mathrm{2}!}\rightarrow{it}\:{is}\:{the}\:{answer} \\ $$$$\frac{\mathrm{8}×\mathrm{7}×\mathrm{6}×\mathrm{5}×\mathrm{3}×\mathrm{2}}{\mathrm{1}}×\frac{\mathrm{4}×\mathrm{3}}{\mathrm{1}}=\mathrm{20160}×\mathrm{6}=\mathrm{120960} \\ $$$${discussion} \\ $$$${assumed}\:{vowel}\:{AAEI}\:{are}\:{Fabiquicked} \\ $$$${permutation}\:{for}\:{AAEI} \\ $$$${is}=\frac{\mathrm{4}}{\mathrm{2}!}\:{denominator}\:\mathrm{2}!\:{for}\:{two}\:{A} \\ $$$${permutation}\:{for}\:\left({AAEI}\right){CHMMSTT} \\ $$$$\frac{\mathrm{8}!}{\mathrm{2}!×\mathrm{2}!}\rightarrow{denominator}\:\mathrm{2}!{for}\:{two}\left(\mathrm{02}\right){M} \\ $$$$\:{anither}\mathrm{2}!\:{for}\:{two}\:{T} \\ $$$${so}\:{answer}=\frac{\mathrm{8}!}{\mathrm{2}!\mathrm{2}!}×\frac{\mathrm{4}!}{\mathrm{2}!} \\ $$$${yes}\:{i}\:{forgot}\:{to}\:{find}\:{probablity}…{MrW}\:{sir}\:{done}\:{it} \\ $$$${pls}\:{check}… \\ $$
Answered by mr W last updated on 08/Mar/19
AAEI =4 letters MMTTHCS =7 letters p=((((8!)/(2!2!))×((4!)/(2!)))/((11!)/(2!2!2!)))=((8!4!)/(11!))=(4/(165))≈2.4%
$${AAEI}\:=\mathrm{4}\:{letters} \\ $$$${MMTTHCS}\:=\mathrm{7}\:{letters} \\ $$$${p}=\frac{\frac{\mathrm{8}!}{\mathrm{2}!\mathrm{2}!}×\frac{\mathrm{4}!}{\mathrm{2}!}}{\frac{\mathrm{11}!}{\mathrm{2}!\mathrm{2}!\mathrm{2}!}}=\frac{\mathrm{8}!\mathrm{4}!}{\mathrm{11}!}=\frac{\mathrm{4}}{\mathrm{165}}\approx\mathrm{2}.\mathrm{4\%} \\ $$
Commented by pieroo last updated on 08/Mar/19
Thanks sir,
$$\mathrm{Thanks}\:\mathrm{sir},\: \\ $$

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