
In how many ways can a vowel, a consonant and a digit be chosen out of 26 letters of English alphabets and 10 digits?
Answer
592.5k+ views
Hint:First we will write what letters are vowel and what are consonants and then we will use the formula of choosing ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$, to find the value of ways a vowel, a consonant and a digit be chosen in each of the cases and then we will multiply them to get the final answer.
Complete step-by-step answer:
Let’s start our solution,
Combination means that in how many ways we can choose from the given number of objects.Now if we have n different objects and from them we need to pick r objects, then the formula is
${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$
Now the vowels are: a, e, i, o, u; which is 5.
The consonant are the rest of English alphabets which is 26 – 5 = 21.
Now for a vowel we have n = 5 and r = 1,
Now substituting these values in ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ we get,
$\begin{align}
& =\dfrac{5!}{1!\left( 5-1 \right)!} \\
& =\dfrac{5\times 4!}{4!} \\
& =5...........(1) \\
\end{align}$
Now for a consonant we have n = 21 and r = 1,
Now substituting these values in ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ we get,
$\begin{align}
& =\dfrac{21!}{1!\left( 21-1 \right)!} \\
& =\dfrac{21\times 20!}{20!} \\
& =21............(2) \\
\end{align}$
Now for digit we have n = 10 and r = 1,
Now substituting these values in ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ we get,
$\begin{align}
& =\dfrac{10!}{1!\left( 10-1 \right)!} \\
& =\dfrac{10\times 9!}{9!} \\
& =10............(3) \\
\end{align}$
Now the final answer will be multiplication of $(1)\times (2)\times (3)$ ,
Hence, we get $5\times 21\times 10=1050$
Hence, the total case is 1050.
Note:We have used the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$, to find all the cases. Here keep in mind that we are not using the formula of permutation because we just have to select, the order in which they are selected doesn’t matter, hence this point must be kept in mind so that there will not be any mistake.
Complete step-by-step answer:
Let’s start our solution,
Combination means that in how many ways we can choose from the given number of objects.Now if we have n different objects and from them we need to pick r objects, then the formula is
${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$
Now the vowels are: a, e, i, o, u; which is 5.
The consonant are the rest of English alphabets which is 26 – 5 = 21.
Now for a vowel we have n = 5 and r = 1,
Now substituting these values in ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ we get,
$\begin{align}
& =\dfrac{5!}{1!\left( 5-1 \right)!} \\
& =\dfrac{5\times 4!}{4!} \\
& =5...........(1) \\
\end{align}$
Now for a consonant we have n = 21 and r = 1,
Now substituting these values in ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ we get,
$\begin{align}
& =\dfrac{21!}{1!\left( 21-1 \right)!} \\
& =\dfrac{21\times 20!}{20!} \\
& =21............(2) \\
\end{align}$
Now for digit we have n = 10 and r = 1,
Now substituting these values in ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ we get,
$\begin{align}
& =\dfrac{10!}{1!\left( 10-1 \right)!} \\
& =\dfrac{10\times 9!}{9!} \\
& =10............(3) \\
\end{align}$
Now the final answer will be multiplication of $(1)\times (2)\times (3)$ ,
Hence, we get $5\times 21\times 10=1050$
Hence, the total case is 1050.
Note:We have used the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$, to find all the cases. Here keep in mind that we are not using the formula of permutation because we just have to select, the order in which they are selected doesn’t matter, hence this point must be kept in mind so that there will not be any mistake.
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