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Find the value of n and r, if \[^n\Pr \]=720 and \[^nCr\]=120
Answer
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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.
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