Chapter 6: Permutations and Combinations

Ans: A number's factorial is the function that multiplies it by each natural number below it. The product of the first $n$ natural numbers is $n$ factorial, which is written as $n!$ and is given by, $n! = n \times \left( {n - 1} \right)!$.

Now, the given expression is $8!$.

According to the definition of the factorial, 

$8! = 8 \times \left( {8 - 1} \right)!$

$8! = 8 \times 7!$

This is further evaluated as,

$ 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ 

$ 8! = 40320 $

Therefore, the value of $8!$ is found to be 40320.

Ans: The product of the first $n$ natural numbers is $n$ factorial, which is written as $n!$ and is given by, $n! = n \times \left( {n - 1} \right)!$.

Now, the given expression is $4! - 3!$.

First, consider and evaluate $4!$.

According to the definition of the factorial, 

$  4! = 4 \times \left( {4 - 1} \right)!$

$  4! = 4 \times 3!$

This is further evaluated as,

$ 4! = 4 \times 3 \times 2 \times 1$ 

$ 4! = 24$

Thus, the value of $4!$ is found to be 24.

Now, consider and evaluate for $3!$.

According to the definition of the factorial, 

$3! = 3 \times \left( {3 - 1} \right)!$

$3! = 3 \times 2! $

This is further evaluated as,

$ 3! = 3 \times 2 \times 1$

$3! = 6$

Thus, the value of $3!$ is found to be 6.

Finally, consider the expression, $4! - 3!$ and evaluate.

$ 4! - 3! = 24 - 6$ 

$4! - 3! = 18$

Therefore, the value of the expression on evaluation is found to be 18.

Ans: The product of the first $n$ natural numbers is $n$ factorial, which is written as $n!$ and is given by, $n! = n \times \left( {n - 1} \right)!$.

First, consider and evaluate $4!$.

According to the definition of the factorial, 

$ 4! = 4 \times \left( {4 - 1} \right)!$ 

$4! = 4 \times 3!$

This is further evaluated as,

$ 4! = 4 \times 3 \times 2 \times 1$ 

$ 4! = 24 $

Thus, the value of $4!$ is found to be 24.

Now, consider and evaluate for $3!$.

According to the definition of the factorial, 

$ 3! = 3 \times \left( {3 - 1} \right)!$

$3! = 3 \times 2!$

This is further evaluated as,

$3! = 3 \times 2 \times 1$

$ 3! = 6$

Thus, the value of $3!$ is found to be 6.

Now, consider the expression, $4! + 3!$ and evaluate.

$ 4! + 3! = 24 + 6$

$ 4! + 3! = 30$

Further, consider $7!$.

According to the definition of the factorial, 

$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$

$7! = 5040$

Therefore, it is observed that $4! + 3! \ne 7!$.

Ans: The product of the first $n$ natural numbers is $n$ factorial, which is written as $n!$ and is given by, $n! = n \times \left( {n - 1} \right)!$.

First consider the expression in the numerator, that is, $8!$.

According to the definition of the factorial, 

$ 8! = 8 \times \left( {8 - 1} \right)!$

$ 8! = 8 \times 7!$

This is further evaluated as,

$8! = 8 \times 7 \times 6!$

Now, consider the expression in the denominator, that is, $6! \times 2!$.

This expression can also be written in a way that is, $6! \times 2 \times 1$.

Consider the expression that is to be determined and evaluated.

$\dfrac{{8!}}{{6! \times 2!}} = \dfrac{{8 \times 7 \times 6!}}{{6! \times 2 \times 1}}$

$\dfrac{{8!}}{{6! \times 2!}} = \dfrac{{8 \times 7}}{2}$

$\dfrac{{8!}}{{6! \times 2!}} = 28$

Therefore, the value of the expression given on evaluation is found to be 28.

Ans: The product of the first $n$ natural numbers is $n$ factorial, which is written as $n!$ and is given by, $n! = n \times \left( {n - 1} \right)!$.

The given expression is $\dfrac{1}{{6!}} + \dfrac{1}{{7!}} = \dfrac{x}{{8!}}$.

Consider making the denominator of the left-hand side of the given expression equal.

$\dfrac{1}{{6!}} + \dfrac{1}{{7 \times 6!}} = \dfrac{x}{{8!}}$

Take out the common terms aside,

$\dfrac{1}{{6!}}\left( {1 + \dfrac{1}{7}} \right) = \dfrac{x}{{8!}}$

$\dfrac{1}{{6!}}\left( {\dfrac{8}{7}} \right) = \dfrac{x}{{8!}}$

Now, rearrange the above equation and solve for $x$.

$x = \dfrac{{8!}}{{6!}}\left( {\dfrac{8}{7}} \right)$

$x = \dfrac{{8 \times 8 \times 7 \times 6!}}{{7 \times 6!}}$

$x = 8 \times 8$

$x = 64$

Therefore, the value of $x$ of the expression on evaluation is found to be 64.

Ans: The expression given is $\dfrac{{n!}}{{\left( {n - r} \right)!}}$.

Substitute 6 for $n$ and 2 for $r$ in the given expression, the expression obtained is,

$\dfrac{{6!}}{{\left( {6 - 2} \right)!}}$

On evaluating the denominator, the expression becomes, $\dfrac{{6!}}{{4!}}$.

Now, consider the expression in the numerator, that is $6!$.

According to the definition of the factorial, 

$6! = 6 \times \left( {6 - 1} \right)!$

$6! = 6 \times 5!$

$6! = 6 \times 5 \times 4!$

Further, consider the expression $\dfrac{{6!}}{{4!}}$.

$\dfrac{{6!}}{{4!}} = \dfrac{{6 \times 5 \times 4!}}{{4!}}$

$\dfrac{{6!}}{{4!}} = 6 \times 5$

$\dfrac{{6!}}{{4!}} = 30$

Therefore, the value of the expression $\dfrac{{n!}}{{\left( {n - r} \right)!}}$, when $n = 6,r = 2$ is found to be 30.

Ans: The expression given is $\dfrac{{n!}}{{\left( {n - r} \right)!}}$.

Substitute 9 for $n$ and 5 for $r$ in the given expression, the expression obtained is,

$\dfrac{{9!}}{{\left( {9 - 5} \right)!}}$

On evaluating the denominator, the expression becomes, $\dfrac{{9!}}{{4!}}$.

Now, consider the expression in the numerator, that is $9!$.

According to the definition of the factorial, 

$9! = 9 \times \left( {9 - 1} \right)! $

$9! = 9 \times 8!$

$9! = 9 \times 8 \times 7 \times 6 \times 5 \times$

Further, consider the expression $\dfrac{{9!}}{{4!}}$.

$\dfrac{{9!}}{{4!}} = \dfrac{{9 \times 8 \times 7 \times 6 \times 5 \times 4!}}{{4!}}$

$\dfrac{{9!}}{{4!}} = 9 \times 8 \times 7 \times 6 \times 5$

$\dfrac{{9!}}{{4!}} = 15120$

Therefore, the value of the expression $\dfrac{{n!}}{{\left( {n - r} \right)!}}$, when $n = 9,r = 5$ is found to be 15120.