Question

There are n different books having m copies each. If the total number of ways of making at least one selection from them is 255 and m-n+1=0, then distance between of point (m, n) from origin is
A) 3
B) 4
C) 5
D) None of these

Answer 1

Hint: In this, we will find the distance between the points (m, n) and origin i.e.(0, 0). To find the distance between the point (m, n) from origin we will first find the value m and n by given condition i.e. the section of at least one book from n different book having m copies and m-n+1=0.

Complete step by step answer:
Given that there are n different point of m copies. Then in the section of at least one book of one type includes that no book is selected of that type. Since there are m copies of each book. Therefore there are m+1 way of selection one type of book. But there are n different book. Hence the number of way of selection of n different book having n copies is${{\left( \text{m+1} \right)}^{n}}$.
This also include the case that no book is selected. Hence the number of at least one selection is =${{\left( \text{m+1} \right)}^{n}}-1$.
Given that the total number of way of selection of at least one book is 255.
$\Rightarrow {{\left( \text{m+1} \right)}^{n}}-1=255$
$\Rightarrow {{\left( \text{m+1} \right)}^{n}}=256....(1)$.
Also, given that m-n+1=0….. (2).
Now we will use trial and elimination methods to find the value of m and n. By checking equation (1) and equation (2)
Since$256={{2}^{8}}={{4}^{4}}={{16}^{2}}.$ satisfies equation (1)
Therefore we have three different cases for m and n
Case 1: m+1 = 2 and n = 8
i.e. m = 1 and n = 8.
\[m-n+1=1-8+1=-6\ne 0\]
Hence m and n does not satisfies the equation (2)
Therefore m and n are not 1 and 8, respectively.
Case 2: m+1=4 and n=4
i.e. m=3 and n=4
\[m-n+1=3-4+1=0\]
m and n satisfies the equation (2)
Hence m and n are 3 and 4, respectively.
Case 2: m+1=16 and n=2
i.e. m=15 and n=2
\[m-n+1=15-2+1=14\ne 0\]
m and n satisfies the equation (2)
Hence m and n are not 15 and 2, respectively.
Hence (m, n) = (3, 4).
The distance of (m, n) = (3, 4) from origin is given below.
$D=\sqrt{{{\left( 3-0 \right)}^{2}}+{{\left( 4-0 \right)}^{2}}}$
\[D=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( 4 \right)}^{2}}}\]
\[D=\sqrt{9+16}=\sqrt{25}\]
D = 5
Hence distance of (m, n) = (3, 4) from origin is 5.

So, the correct answer is “Option C”.

Note: In this problem students should keep in mind the number of ways of selection also includes no selection. Therefore there are ${{\left( \text{m+1} \right)}^{n}}-1$ way of selection and the distance between two points ($x_1, y_1$) and ($x_2, y_2$) is
$D=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}.$