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The-number-of-words-can-be-made-formed-using-all-letters-in-the-word-amutasia-by-not-containing-the-three-vowels-side-by-side-is-




Question Number 77777 by jagoll last updated on 10/Jan/20
The number  of words can be made formed  using all letters in the word  ′amutasia′ by not containing  the three vowels side by side is ?
$${The}\:{number} \\ $$$${of}\:{words}\:{can}\:{be}\:{made}\:{formed} \\ $$$${using}\:{all}\:{letters}\:{in}\:{the}\:{word} \\ $$$$'{amutasia}'\:{by}\:{not}\:{containing} \\ $$$${the}\:{three}\:{vowels}\:{side}\:{by}\:{side}\:{is}\:? \\ $$
Answered by mr W last updated on 10/Jan/20
AMUTASIA contains  3× A  1× I  1× U  1× M  1× S  1× T  totally 8 letters    total number of words formed using  all letters:  ((8!)/(3!))=6720  among them number of words containing  the three vowels side by side:  ((3×6!3!)/(2!))=6480    ⇒requested number: 6720−6480=240
$${AMUTASIA}\:{contains} \\ $$$$\mathrm{3}×\:{A} \\ $$$$\mathrm{1}×\:{I} \\ $$$$\mathrm{1}×\:{U} \\ $$$$\mathrm{1}×\:{M} \\ $$$$\mathrm{1}×\:{S} \\ $$$$\mathrm{1}×\:{T} \\ $$$${totally}\:\mathrm{8}\:{letters} \\ $$$$ \\ $$$${total}\:{number}\:{of}\:{words}\:{formed}\:{using} \\ $$$${all}\:{letters}: \\ $$$$\frac{\mathrm{8}!}{\mathrm{3}!}=\mathrm{6720} \\ $$$${among}\:{them}\:{number}\:{of}\:{words}\:{containing} \\ $$$${the}\:{three}\:{vowels}\:{side}\:{by}\:{side}: \\ $$$$\frac{\mathrm{3}×\mathrm{6}!\mathrm{3}!}{\mathrm{2}!}=\mathrm{6480} \\ $$$$ \\ $$$$\Rightarrow{requested}\:{number}:\:\mathrm{6720}−\mathrm{6480}=\mathrm{240} \\ $$
Commented by jagoll last updated on 10/Jan/20
sir why ((3×6!3!)/(2! )) sir?please give example
$${sir}\:{why}\:\frac{\mathrm{3}×\mathrm{6}!\mathrm{3}!}{\mathrm{2}!\:}\:{sir}?{please}\:{give}\:{example} \\ $$
Commented by mr W last updated on 10/Jan/20
in fact the question is not exact.  the three vowols should not be side by side.  but are SIAAUMAT and SIAAAUMT  valid? for me they are valid. side by  side means for me IAU, but not IAAU.
$${in}\:{fact}\:{the}\:{question}\:{is}\:{not}\:{exact}. \\ $$$${the}\:{three}\:{vowols}\:{should}\:{not}\:{be}\:{side}\:{by}\:{side}. \\ $$$${but}\:{are}\:{SIAAUMAT}\:{and}\:{SIAAAUMT} \\ $$$${valid}?\:{for}\:{me}\:{they}\:{are}\:{valid}.\:{side}\:{by} \\ $$$${side}\:{means}\:{for}\:{me}\:{IAU},\:{but}\:{not}\:{IAAU}. \\ $$
Commented by jagoll last updated on 10/Jan/20
valid sir
$${valid}\:{sir} \\ $$
Commented by mr W last updated on 10/Jan/20
then the answer is 240. is this correct?
$${then}\:{the}\:{answer}\:{is}\:\mathrm{240}.\:{is}\:{this}\:{correct}? \\ $$
Commented by jagoll last updated on 10/Jan/20
the problem is not multiple   choice sir. i still don′t   understand the formula  reduction ((3×6!3!)/(2! ))
$${the}\:{problem}\:{is}\:{not}\:{multiple}\: \\ $$$${choice}\:{sir}.\:{i}\:{still}\:{don}'{t}\: \\ $$$${understand}\:{the}\:{formula} \\ $$$${reduction}\:\frac{\mathrm{3}×\mathrm{6}!\mathrm{3}!}{\mathrm{2}!\:}\: \\ $$
Commented by jagoll last updated on 10/Jan/20
the first 3 number are many types  of vowels . why is divided 2! ?
$${the}\:{first}\:\mathrm{3}\:{number}\:{are}\:{many}\:{types} \\ $$$${of}\:{vowels}\:.\:{why}\:{is}\:{divided}\:\mathrm{2}!\:? \\ $$
Commented by malwaan last updated on 10/Jan/20
with repeating letters or not?
$${with}\:{repeating}\:{letters}\:{or}\:{not}? \\ $$
Commented by jagoll last updated on 10/Jan/20
not sir
$${not}\:{sir} \\ $$
Commented by mr W last updated on 10/Jan/20
to jagoll sir:  image you mark the three letters A with  a pen as A_1 ,A_2 ,A_3  and treat them as  different letters. certainly they are  in fact identical.  then we select three vowols, e.g. A_1 ,I,U,  as a group, say [A_1 IU], as if they were   a single letter. then we arrange the  6 “letters”: [A_1 IU],A_2 ,A_3 ,M,S,T.  there are 6! ways to arrange them.  but A_2  and A_3  are in fact identical,  therefore there are in fact only  ((6!)/(2!)) different ways. on the other side,  we can arrange the three vowols A_1 ,I,U  inside the group in 3! ways. therefore we have  totally ((6!3!)/(2!)) ways when we take A_1  as  vowol. since we can also select A_2  or  A_3  as vowol, the number of all possibilties  is ((3×6!3!)/(2!)).  clear?
$${to}\:{jagoll}\:{sir}: \\ $$$${image}\:{you}\:{mark}\:{the}\:{three}\:{letters}\:{A}\:{with} \\ $$$${a}\:{pen}\:{as}\:{A}_{\mathrm{1}} ,{A}_{\mathrm{2}} ,{A}_{\mathrm{3}} \:{and}\:{treat}\:{them}\:{as} \\ $$$${different}\:{letters}.\:{certainly}\:{they}\:{are} \\ $$$${in}\:{fact}\:{identical}. \\ $$$${then}\:{we}\:{select}\:{three}\:{vowols},\:{e}.{g}.\:{A}_{\mathrm{1}} ,{I},{U}, \\ $$$${as}\:{a}\:{group},\:{say}\:\left[{A}_{\mathrm{1}} {IU}\right],\:{as}\:{if}\:{they}\:{were}\: \\ $$$${a}\:{single}\:{letter}.\:{then}\:{we}\:{arrange}\:{the} \\ $$$$\mathrm{6}\:“{letters}'':\:\left[{A}_{\mathrm{1}} {IU}\right],{A}_{\mathrm{2}} ,{A}_{\mathrm{3}} ,{M},{S},{T}. \\ $$$${there}\:{are}\:\mathrm{6}!\:{ways}\:{to}\:{arrange}\:{them}. \\ $$$${but}\:{A}_{\mathrm{2}} \:{and}\:{A}_{\mathrm{3}} \:{are}\:{in}\:{fact}\:{identical}, \\ $$$${therefore}\:{there}\:{are}\:{in}\:{fact}\:{only} \\ $$$$\frac{\mathrm{6}!}{\mathrm{2}!}\:{different}\:{ways}.\:{on}\:{the}\:{other}\:{side}, \\ $$$${we}\:{can}\:{arrange}\:{the}\:{three}\:{vowols}\:{A}_{\mathrm{1}} ,{I},{U} \\ $$$${inside}\:{the}\:{group}\:{in}\:\mathrm{3}!\:{ways}.\:{therefore}\:{we}\:{have} \\ $$$${totally}\:\frac{\mathrm{6}!\mathrm{3}!}{\mathrm{2}!}\:{ways}\:{when}\:{we}\:{take}\:{A}_{\mathrm{1}} \:{as} \\ $$$${vowol}.\:{since}\:{we}\:{can}\:{also}\:{select}\:{A}_{\mathrm{2}} \:{or} \\ $$$${A}_{\mathrm{3}} \:{as}\:{vowol},\:{the}\:{number}\:{of}\:{all}\:{possibilties} \\ $$$${is}\:\frac{\mathrm{3}×\mathrm{6}!\mathrm{3}!}{\mathrm{2}!}. \\ $$$${clear}? \\ $$
Commented by jagoll last updated on 10/Jan/20
yes sir. thanks you very much
$${yes}\:{sir}.\:{thanks}\:{you}\:{very}\:{much} \\ $$
Commented by malwaan last updated on 10/Jan/20
great sir Mr W  thank you very much
$$\boldsymbol{{great}}\:\boldsymbol{{sir}}\:\boldsymbol{{Mr}}\:\boldsymbol{{W}} \\ $$$$\boldsymbol{{thank}}\:\boldsymbol{{you}}\:\boldsymbol{{very}}\:\boldsymbol{{much}} \\ $$$$ \\ $$

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