Counting Problem
\(\large \color{black}{\begin{align} & \normalsize \text{How many words of 11 letters could be formed with } \hspace{.33em}\\~\\ & \normalsize \text{all the vowels present in even places, using all the } \hspace{.33em}\\~\\ & \normalsize \text{letters of the alphabet ?(without repetition) } \hspace{.33em}\\~\\ \end{align}}\)
what are your thoughts?
\(\large \color{black}{\begin{align} & a.)\ ^{21}P_{6}\times 5! \hspace{.33em}\\~\\ & b.)\ 21! \hspace{.33em}\\~\\ & c.)\ ^{21}P_{5}\times 5! \hspace{.33em}\\~\\ & d.)\ ^{26}P_{8} \hspace{.33em}\\~\\ \end{align}}\)
|dw:1440438880520:dw|
i m confused on how to think here
those 5 places are blocked with 5 vowels, which can be arranged among themselves in how many ways??
6 letters should be vowels and 5 letters non vowels
how many other letters are left? in how many ways can you arrange those other letters in 6 places?
5 vowels, at places: 2,4,6,8,10
\(\large \color{black}{\begin{align} & ^{6}P_{5}\ ways \hspace{.33em}\\~\\ \end{align}}\)
iis it correct
sorry for asking many questions at one time. lets just concentrate on 5 places are blocked with 5 vowels, which can be arranged among themselves in how many ways??
5! ways
good! now forget about those 5 places, they are done only 6 more places to go, and how many options do we have? hint : consonants
there are 6 places and 21 consonants remaining
and how can we arrange that?
\(\large \color{black}{\begin{align} & ^{21}P_{6}\ ways \hspace{.33em}\\~\\ \end{align}}\)
\(\huge \checkmark \)
any more doubts? :)
yes
does u mean 1st option is correct
ok thanks
yes. clear any doubts that you have in your mind ...
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