A man has 7 trousers and 10 shirts. How many different outfits can he wear?
Answer
626.7k+ views
Hint: To wear an outfit he must choose 1 trouser from 7 and 1 shirt from 10 shirts. We will use the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ for choosing r things from n given things. Hence, we will use this to find the total number of ways he can choose 1 trouser and 1 shirt.
Complete step-by-step answer:
Let’s start our solution.
We have to choose 1 trouser from 7 and 1 shirt from 10 shirts.
We will use the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ for choosing r things from n given things.
Let’s first choose trousers.
Now for trousers we have n = 7 and r = 1.
Substituting the value in ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ we get,
$\begin{align}
& ={}^{7}{{C}_{1}} \\
& =\dfrac{7!}{1!\left( 7-1 \right)!} \\
& =\dfrac{7\times 6!}{6!} \\
& =7 \\
\end{align}$
Hence, the number of ways he can choose a trouser is 7.
Let’s choose a shirt.
Now for trousers we have n = 10 and r = 1.
Substituting the value in ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ we get,
$\begin{align}
& ={}^{10}{{C}_{1}} \\
& =\dfrac{10!}{1!\left( 10-1 \right)!} \\
& =\dfrac{10\times 9!}{9!} \\
& =10 \\
\end{align}$
Hence, the number of ways he can choose a shirt is 10.
Now we will multiply the number of ways of choosing shirts to the trouser to get the total number of outfits.
Now the number of different he can wear is $7\times 10=70$
Hence, the answer is 70.
Note: The student should understand how we got 70, we can see that for each trouser there are 10 different shirts and so for 1 trouser we have 10 different outfits, so similarly we can say that for 7 different trousers we will have 70 different outfits.
Complete step-by-step answer:
Let’s start our solution.
We have to choose 1 trouser from 7 and 1 shirt from 10 shirts.
We will use the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ for choosing r things from n given things.
Let’s first choose trousers.
Now for trousers we have n = 7 and r = 1.
Substituting the value in ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ we get,
$\begin{align}
& ={}^{7}{{C}_{1}} \\
& =\dfrac{7!}{1!\left( 7-1 \right)!} \\
& =\dfrac{7\times 6!}{6!} \\
& =7 \\
\end{align}$
Hence, the number of ways he can choose a trouser is 7.
Let’s choose a shirt.
Now for trousers we have n = 10 and r = 1.
Substituting the value in ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ we get,
$\begin{align}
& ={}^{10}{{C}_{1}} \\
& =\dfrac{10!}{1!\left( 10-1 \right)!} \\
& =\dfrac{10\times 9!}{9!} \\
& =10 \\
\end{align}$
Hence, the number of ways he can choose a shirt is 10.
Now we will multiply the number of ways of choosing shirts to the trouser to get the total number of outfits.
Now the number of different he can wear is $7\times 10=70$
Hence, the answer is 70.
Note: The student should understand how we got 70, we can see that for each trouser there are 10 different shirts and so for 1 trouser we have 10 different outfits, so similarly we can say that for 7 different trousers we will have 70 different outfits.
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