Answer
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Hint: We have given that there are 5 books on Mathematics and 6 books on Physics in a book shop. The possible ways in which we can buy a Mathematics and a Physics book are selecting a Mathematics book from 5 Mathematics books using a combinatorial approach and then multiply the selection of a Physics book from 6 Physics books. After that, the possible ways to buy either a Mathematics book or a physics book are selecting a Mathematics book from 5 Mathematics books using a combinatorial approach and then add the result of this selection to the selection of a Physics book from 6 Physics books.
Complete step-by-step solution:
In a book shop, there are 5 books on Mathematics and 6 books on Physics. And we have to find possible ways to buy:
In the first part, we have to buy a Mathematics and a Physics book which we are going to calculate as follows:
By selecting a Mathematics book from 5 Mathematics books using the combinatorial approach we get,
${}^{5}{{C}_{1}}$
Now, multiplying the above by the selection of a Physics book from 6 Physics books we get,
${}^{5}{{C}_{1}}.{}^{6}{{C}_{1}}$
We know that,
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Using the above relation to find the value of ${}^{5}{{C}_{1}}.{}^{6}{{C}_{1}}$ we get,
$\begin{align}
& \dfrac{5!}{1!\left( 5-1 \right)!}\left( \dfrac{6!}{1!\left( 6-1 \right)!} \right) \\
& =\dfrac{5!}{1!\left( 4 \right)!}\left( \dfrac{6!}{1!\left( 5 \right)!} \right) \\
\end{align}$
Now, in the above expression, we can write 5! as 5.4! and 6! as 6.5!
$\dfrac{5.4!}{1!\left( 4 \right)!}\left( \dfrac{6.5!}{1!\left( 5 \right)!} \right)$
In the above expression, $5!$ and $4!$ will be cancelled out from the numerator and denominator.
$\dfrac{5}{1!}\left( \dfrac{6}{1!} \right)$
And we know that the value of $1!=1$ so using this relation in the above expression we get,
$\begin{align}
& 5.6 \\
& =30 \\
\end{align}$
Hence, the possible number of ways to buy a Mathematics and a Physics book is 30.
Now, we are going to solve the part (ii).
In this part, we have to find the possible number of ways to buy either a Mathematics book or a Physics book which we are going to do by selecting a Mathematics book from 5 books first.
${}^{5}{{C}_{1}}$
And also we are selecting a Physics book from 6 Physics books we get,
${}^{6}{{C}_{1}}$
Now, adding both the selection of a Mathematics and a Physics book which we have just shown we get,
${}^{5}{{C}_{1}}+{}^{6}{{C}_{1}}$
We have already solved the value of ${}^{5}{{C}_{1}}\And {}^{6}{{C}_{1}}$ as 5 and 6 in the previous part so substituting these values in the above expression we get,
$\begin{align}
& 5+6 \\
& =11 \\
\end{align}$
Hence, we found the possible number of ways to buy either a Mathematics book or a Physics book as 11.
Note: You can deduce information from this problem is that whenever you have given the expression “and” which is shown in the first part of the problem that we have to buy a Mathematics and a Physics book then we “multiply” as we did in the above solution by multiplying the ways of selecting a Mathematics book with the ways of selecting a Physics book.
Also, if you have given “either-or” then we do the “addition” of the possible ways which we did in the second part of this problem by adding the ways of selecting a Mathematics book with the ways of selecting a Physics book.
Complete step-by-step solution:
In a book shop, there are 5 books on Mathematics and 6 books on Physics. And we have to find possible ways to buy:
In the first part, we have to buy a Mathematics and a Physics book which we are going to calculate as follows:
By selecting a Mathematics book from 5 Mathematics books using the combinatorial approach we get,
${}^{5}{{C}_{1}}$
Now, multiplying the above by the selection of a Physics book from 6 Physics books we get,
${}^{5}{{C}_{1}}.{}^{6}{{C}_{1}}$
We know that,
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Using the above relation to find the value of ${}^{5}{{C}_{1}}.{}^{6}{{C}_{1}}$ we get,
$\begin{align}
& \dfrac{5!}{1!\left( 5-1 \right)!}\left( \dfrac{6!}{1!\left( 6-1 \right)!} \right) \\
& =\dfrac{5!}{1!\left( 4 \right)!}\left( \dfrac{6!}{1!\left( 5 \right)!} \right) \\
\end{align}$
Now, in the above expression, we can write 5! as 5.4! and 6! as 6.5!
$\dfrac{5.4!}{1!\left( 4 \right)!}\left( \dfrac{6.5!}{1!\left( 5 \right)!} \right)$
In the above expression, $5!$ and $4!$ will be cancelled out from the numerator and denominator.
$\dfrac{5}{1!}\left( \dfrac{6}{1!} \right)$
And we know that the value of $1!=1$ so using this relation in the above expression we get,
$\begin{align}
& 5.6 \\
& =30 \\
\end{align}$
Hence, the possible number of ways to buy a Mathematics and a Physics book is 30.
Now, we are going to solve the part (ii).
In this part, we have to find the possible number of ways to buy either a Mathematics book or a Physics book which we are going to do by selecting a Mathematics book from 5 books first.
${}^{5}{{C}_{1}}$
And also we are selecting a Physics book from 6 Physics books we get,
${}^{6}{{C}_{1}}$
Now, adding both the selection of a Mathematics and a Physics book which we have just shown we get,
${}^{5}{{C}_{1}}+{}^{6}{{C}_{1}}$
We have already solved the value of ${}^{5}{{C}_{1}}\And {}^{6}{{C}_{1}}$ as 5 and 6 in the previous part so substituting these values in the above expression we get,
$\begin{align}
& 5+6 \\
& =11 \\
\end{align}$
Hence, we found the possible number of ways to buy either a Mathematics book or a Physics book as 11.
Note: You can deduce information from this problem is that whenever you have given the expression “and” which is shown in the first part of the problem that we have to buy a Mathematics and a Physics book then we “multiply” as we did in the above solution by multiplying the ways of selecting a Mathematics book with the ways of selecting a Physics book.
Also, if you have given “either-or” then we do the “addition” of the possible ways which we did in the second part of this problem by adding the ways of selecting a Mathematics book with the ways of selecting a Physics book.