Find the value of n and r, if \[^n\Pr \]=720 and \[^nCr\]=120
Answer
Verified
483.3k+ views
Hint: A permutation is defined as an arrangement in a definite order of a number of objects taken some or all at a time. The convenient expression to denote permutation is defined as
The permutation formula is given by,
\[^n\Pr = \dfrac{{n!}}{{\left( {n - r} \right)!}};0 \leqslant r \leqslant n\]
Where the symbol denotes the factorial which means that the product of all the integer less than or equal to n but it should be greater than or equal to 1.
Combination- the combination is a selection of a part of a set of objects or selection of all objects when the order does not matter. Therefore, the number of combinations of n objects taken r at a time and the combination formula is given by,
\[^nCr = \dfrac{{n(n - 1)(n - 2).....(n - r + 1)}}{{r!}}\]
\[ = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
\[^nCr = \dfrac{{^n\Pr }}{{r!}}\]
Therefore,
Complete step by step answer:
Given, \[^n\Pr = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
Which is equal to \[720\]
\[\dfrac{{n!}}{{\left( {n - r} \right)!}} = 720\]
As we know the relation between permutation and combination
\[^n\Pr = r{!^n}Cr\]
\[\dfrac{{^n\Pr }}{{^nCr}} = r!.......1.\]
Also given in the question \[^nCr = 120\]
Putting the value of \[^n\Pr \] and \[^nCr\] in equation 1.
\[\dfrac{{720}}{{120}} = r!\]
\[r! = 6\]
\[r! = 3 \times 2 \times 1\]
\[r = 3\]
Now, \[^n\operatorname{P} 3 = 720\]
We can write this
\[n(n - 1)(n - 2) = 720\]
\[n(n - 1)(n - 2) = 10 \times 9 \times 8\]
From this we get
\[n = 10\]
Hence the value of \[n\] and \[r\] are \[10\] and \[3\] respectively.
Note: The relation between permutation and combination-
\[^n\Pr { = ^n}Cr.r!\] if
\[0 < r \leqslant n\]
\[^nCr{ + ^n}Cr - 1{ = ^{n + 1}}Cr\]
The fundamental principle of counting
Multiplication principal
Suppose an operation
The fundamental principle of counting –
Multiplication principle: suppose an operation A can be performed in m ways and associated with each way of performing another operation B can be performed in n ways, then the total number of performances of two operations in the given order is \[m \times n\] ways. This can be extended to any finite number of operations.
Addition principle: if an operation A can be performed in m ways and another operation S, which is independent of A, can be performed in \[m + n\] ways. This can be extended to any finite number of exclusive events.
The permutation formula is given by,
\[^n\Pr = \dfrac{{n!}}{{\left( {n - r} \right)!}};0 \leqslant r \leqslant n\]
Where the symbol denotes the factorial which means that the product of all the integer less than or equal to n but it should be greater than or equal to 1.
Combination- the combination is a selection of a part of a set of objects or selection of all objects when the order does not matter. Therefore, the number of combinations of n objects taken r at a time and the combination formula is given by,
\[^nCr = \dfrac{{n(n - 1)(n - 2).....(n - r + 1)}}{{r!}}\]
\[ = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
\[^nCr = \dfrac{{^n\Pr }}{{r!}}\]
Therefore,
Complete step by step answer:
Given, \[^n\Pr = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
Which is equal to \[720\]
\[\dfrac{{n!}}{{\left( {n - r} \right)!}} = 720\]
As we know the relation between permutation and combination
\[^n\Pr = r{!^n}Cr\]
\[\dfrac{{^n\Pr }}{{^nCr}} = r!.......1.\]
Also given in the question \[^nCr = 120\]
Putting the value of \[^n\Pr \] and \[^nCr\] in equation 1.
\[\dfrac{{720}}{{120}} = r!\]
\[r! = 6\]
\[r! = 3 \times 2 \times 1\]
\[r = 3\]
Now, \[^n\operatorname{P} 3 = 720\]
We can write this
\[n(n - 1)(n - 2) = 720\]
\[n(n - 1)(n - 2) = 10 \times 9 \times 8\]
From this we get
\[n = 10\]
Hence the value of \[n\] and \[r\] are \[10\] and \[3\] respectively.
Note: The relation between permutation and combination-
\[^n\Pr { = ^n}Cr.r!\] if
\[0 < r \leqslant n\]
\[^nCr{ + ^n}Cr - 1{ = ^{n + 1}}Cr\]
The fundamental principle of counting
Multiplication principal
Suppose an operation
The fundamental principle of counting –
Multiplication principle: suppose an operation A can be performed in m ways and associated with each way of performing another operation B can be performed in n ways, then the total number of performances of two operations in the given order is \[m \times n\] ways. This can be extended to any finite number of operations.
Addition principle: if an operation A can be performed in m ways and another operation S, which is independent of A, can be performed in \[m + n\] ways. This can be extended to any finite number of exclusive events.
Recently Updated Pages
What percentage of the area in India is covered by class 10 social science CBSE
The area of a 6m wide road outside a garden in all class 10 maths CBSE
What is the electric flux through a cube of side 1 class 10 physics CBSE
If one root of x2 x k 0 maybe the square of the other class 10 maths CBSE
The radius and height of a cylinder are in the ratio class 10 maths CBSE
An almirah is sold for 5400 Rs after allowing a discount class 10 maths CBSE
Trending doubts
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Why is there a time difference of about 5 hours between class 10 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE
Explain the Treaty of Vienna of 1815 class 10 social science CBSE
Write an application to the principal requesting five class 10 english CBSE