
If 5 vowels and 6 consonants are given, then how many 6 letter words can be formed with 3 vowels and 3 consonants?
Answer
486.6k+ views
Hint: First we will find the number of ways to choose vowels and consonants separately by using the formula of combination, which is given by
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
Where, $n=$ number of items/objects
And $r=$ number of items/objects being chosen at a time
Then, we find the number of ways to choose both vowels and consonants. Then, find the number of ways to arrange them to form 6 letter words. Then, multiply the obtained numbers to get the desired result.
Complete step by step answer:
We have given 5 vowels and 6 consonants.
Then, we have to find how many 6 letter words can be formed with 3 vowels and 3 consonants.
Now, we need to choose 3 vowels from the given 5 vowels. So, the number of ways to choose vowels will be
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
\[\begin{align}
&\Rightarrow {}^{5}{{C}_{3}}=\dfrac{5!}{3!\left( 5-3 \right)!} \\
&\Rightarrow {}^{5}{{C}_{3}}=\dfrac{5\times 4\times 3!}{3!\left( 2 \right)!} \\
&\Rightarrow {}^{5}{{C}_{3}}=\dfrac{5\times 4}{2\times 1} \\
&\Rightarrow {}^{5}{{C}_{3}}=\dfrac{20}{2} \\
&\Rightarrow {}^{5}{{C}_{3}}=10 \\
\end{align}\]
Now, we have to find the number of ways to choose 3 consonants from the given 6 consonants, we get
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
\[\begin{align}
&\Rightarrow {}^{6}{{C}_{3}}=\dfrac{6!}{3!\left( 6-3 \right)!} \\
& \Rightarrow {}^{6}{{C}_{3}}=\dfrac{6\times 5\times 4\times 3!}{3!\left( 3 \right)!} \\
& \Rightarrow {}^{6}{{C}_{3}}=\dfrac{6\times 5\times 4}{3\times 2\times 1} \\
&\Rightarrow {}^{6}{{C}_{3}}=\dfrac{120}{6} \\
&\Rightarrow {}^{6}{{C}_{3}}=20 \\
\end{align}\]
Now, we have to find the number of ways to select both vowels and consonants.
We have \[{}^{5}{{C}_{3}}=10\] and \[{}^{6}{{C}_{3}}=20\], so number of ways to choose both will be
\[\begin{align}
& {}^{5}{{C}_{3}}\times {}^{6}{{C}_{3}}=20\times 10 \\
& {}^{5}{{C}_{3}}\times {}^{6}{{C}_{3}}=200 \\
\end{align}\]
Now, we have to form 6 letter words by arranging this vowels and consonants. So, 6 letters are arranged in $6!$ ways.
So, the total number of words that can be formed will be
$\begin{align}
& 6!\times 200 \\
& =6\times 5\times 4\times 3\times 2\times 1\times 200 \\
& =144000 \\
\end{align}$
So, total $144000$ words can be formed.
Note: There is a possibility that students may forget to arrange 6 letters to form a word and give the answer as $200$, which is an incorrect answer. Each arrangement of 6 letters gives different words so it is necessary to arrange them within themselves. Also, there is a difference between permutation and combination. Combination means only choosing while permutation means first choosing then arranging.
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
Where, $n=$ number of items/objects
And $r=$ number of items/objects being chosen at a time
Then, we find the number of ways to choose both vowels and consonants. Then, find the number of ways to arrange them to form 6 letter words. Then, multiply the obtained numbers to get the desired result.
Complete step by step answer:
We have given 5 vowels and 6 consonants.
Then, we have to find how many 6 letter words can be formed with 3 vowels and 3 consonants.
Now, we need to choose 3 vowels from the given 5 vowels. So, the number of ways to choose vowels will be
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
\[\begin{align}
&\Rightarrow {}^{5}{{C}_{3}}=\dfrac{5!}{3!\left( 5-3 \right)!} \\
&\Rightarrow {}^{5}{{C}_{3}}=\dfrac{5\times 4\times 3!}{3!\left( 2 \right)!} \\
&\Rightarrow {}^{5}{{C}_{3}}=\dfrac{5\times 4}{2\times 1} \\
&\Rightarrow {}^{5}{{C}_{3}}=\dfrac{20}{2} \\
&\Rightarrow {}^{5}{{C}_{3}}=10 \\
\end{align}\]
Now, we have to find the number of ways to choose 3 consonants from the given 6 consonants, we get
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
\[\begin{align}
&\Rightarrow {}^{6}{{C}_{3}}=\dfrac{6!}{3!\left( 6-3 \right)!} \\
& \Rightarrow {}^{6}{{C}_{3}}=\dfrac{6\times 5\times 4\times 3!}{3!\left( 3 \right)!} \\
& \Rightarrow {}^{6}{{C}_{3}}=\dfrac{6\times 5\times 4}{3\times 2\times 1} \\
&\Rightarrow {}^{6}{{C}_{3}}=\dfrac{120}{6} \\
&\Rightarrow {}^{6}{{C}_{3}}=20 \\
\end{align}\]
Now, we have to find the number of ways to select both vowels and consonants.
We have \[{}^{5}{{C}_{3}}=10\] and \[{}^{6}{{C}_{3}}=20\], so number of ways to choose both will be
\[\begin{align}
& {}^{5}{{C}_{3}}\times {}^{6}{{C}_{3}}=20\times 10 \\
& {}^{5}{{C}_{3}}\times {}^{6}{{C}_{3}}=200 \\
\end{align}\]
Now, we have to form 6 letter words by arranging this vowels and consonants. So, 6 letters are arranged in $6!$ ways.
So, the total number of words that can be formed will be
$\begin{align}
& 6!\times 200 \\
& =6\times 5\times 4\times 3\times 2\times 1\times 200 \\
& =144000 \\
\end{align}$
So, total $144000$ words can be formed.
Note: There is a possibility that students may forget to arrange 6 letters to form a word and give the answer as $200$, which is an incorrect answer. Each arrangement of 6 letters gives different words so it is necessary to arrange them within themselves. Also, there is a difference between permutation and combination. Combination means only choosing while permutation means first choosing then arranging.
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