Rd sharma solution of class 11 chapter 16 permutation

EXERCISE 16.1

QUESTION 1

Compute :

(i) \dfrac{30 !}{28 !}

Sol :

=\dfrac{30 \times 29 \times 28 !}{28 !} [\because n !=n(n-1) !]

=30 \times 29

= 870

 

(ii) \dfrac{11 !-10 !}{9 !}

Sol :

=\dfrac{11 \times 10 \times 9 !-10 \times 9 !}{9 !} [\because n !=n(n-1) !]

=\dfrac{9 !(110-10)}{9 !}

= 100

 

(iii) L.C.M. (6 !, 7 !, 8 !)

Sol :

n !=n(n-1) !

Therefore, (6 !, 7 ! and 8 !) can be rewritten as:

8 !=8 \times 7 \times 6 !

7 !=7 \times 6 !

6 !=6 !

LCM of (6 !, 7 ! and 8 !) =L C M[8 \times 7 \times 6 !, 7 \times 6 !, 6 !] =8 \times 7 \times 6 !=8 !

 

QUESTION 2

Prove that \dfrac{1}{9 !}+\dfrac{1}{10 !}+\dfrac{1}{11 !}=\dfrac{122}{11 !}

Sol :

\mathrm{LHS}=\dfrac{1}{9 !}+\dfrac{1}{10 !}+\dfrac{1}{11 !}

=\dfrac{1}{9 !}+\dfrac{1}{10 \times 9 !}+\dfrac{1}{11 \times 10 \times 9 !}

=\dfrac{110+11+1}{11 \times 10 \times 9 !}

=\dfrac{122}{11 !}=\mathrm{RHS}

Hence , proved

 

QUESTION 3

Find x in each of the following:

(i) \dfrac{1}{4 !}+\dfrac{1}{5 !}=\dfrac{x}{6 !}

Sol :

\Rightarrow \dfrac{1}{4 !}+\dfrac{1}{5(4 !)}=\dfrac{x}{6 !}

\Rightarrow \dfrac{5+1}{5(4 !)}=\dfrac{x}{6 !}

\Rightarrow \dfrac{5+1}{5(4 !)}=\dfrac{x}{6 !}

\Rightarrow \dfrac{6}{5 !}=\dfrac{x}{6 \times 5 !}

\Rightarrow x=36

 

(ii) \dfrac{x}{10 !}=\dfrac{1}{8 !}+\dfrac{1}{9 !}

Sol :

\Rightarrow \dfrac{x}{10 !}=\dfrac{1}{8 !}+\dfrac{1}{9(8 !)}

\Rightarrow \dfrac{x}{10 !}=\dfrac{9+1}{9(8 !)}

\Rightarrow \dfrac{x}{10 !}=\dfrac{10}{9 !}

\Rightarrow \dfrac{x}{10 \times 9 !}=\dfrac{10}{9 !}

\Rightarrow x=100

 

(iii) \dfrac{1}{6 !}+\dfrac{1}{7 !}=\dfrac{x}{8 !}

Sol :

\Rightarrow \dfrac{1}{6 !}+\dfrac{1}{7(6 !)}=\dfrac{x}{8 !}

\Rightarrow \dfrac{7+1}{7(6 !)}=\dfrac{x}{8 !}

\Rightarrow \quad \dfrac{8}{7 !}=\dfrac{x}{8 !}

\Rightarrow \quad \dfrac{8}{7 !}=\dfrac{x}{8 \times 7 !}

\Rightarrow x=64

 

QUESTION 4

Convert the following products into factorials:

(i) 5\cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10

Sol :

=\dfrac{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10}{1 \times 2 \times 3 \times 4}

=\dfrac{10 !}{4 !}

 

(ii) 3\cdot 6 \cdot 9 \cdot 12 \cdot 15 \cdot 18

Sol :

=(3 \times 1) \times(3 \times 2) \times(3 \times 3) \times(3 \times 4) \times(3 \times 5) \times(3 \times 6)

=3^{6}(1 \times 2 \times 3 \times 4 \times 5 \times 6)

=3^{6}(6 !)

 

(iii) (n+1)(n+2)(n+3) \ldots(2 n)

Sol :

=\dfrac{(1)(2)(3) \dots(n)(n+1)(n+2)(n+3) \ldots(2 n)}{(1)(2)(3) \dots(n)}

=\dfrac{(2 n)}{n !}

 

(iv) 1\cdot 3 \cdot 5 \cdot 7 \cdot 9 \ldots(2 n-1)

Sol :

=\dfrac{[(1)(3)(5)\dots (2n-1)][(2)(4)(6)\dots (2n)]}{[(2)(4)(6)\dots(2n)]}

=\dfrac{[(1)(2)(3)(4)(5) \ldots \ldots \ldots(2 n-1)(2 n)]}{2^{n}[(1)(2)(3) \ldots \ldots(n)]}

=\dfrac{(2 n) !}{2^{n} n !}

 

QUESTION 5

Which of the following are true :

(i) (2+3)!=2!+3!

Sol :

\text{ L H S }=(2+3) !

=5 !

=120

\text{ R H S }=2 !+3 !

=2+6

= 8

Since \mathrm{LHS} \neq \mathrm{RHS}

Thus , (i) is false

 

(ii) (2\times 3)!=2!\times 3!

Sol :

\text{ L H S }=(2 \times 3) !

=6 !

= 720

\text{ R H S }=2 ! \times 3 !

=2 \times 6

= 12

\mathrm{LHS} \neq \mathrm{RHS}

Thus , (ii) is false

 

QUESTION 6

Prove that: n !(n+2)=n !+(n+1) !

Sol :

\mathrm{RHS}=n !+(n+1) !

=n !+(n+1)(n !)

=n !(1+n+1)

=n !(n+2)=\text{ L H S }

Hence proved

 

QUESTION 7

If (n+2) !=60[(n-1) !], find n

Sol :

(n+2) !=60[(n-1) !]

\Rightarrow(n+2) \times(n+1) \times(n) \times(n-1) !=60[(n-1) !]

\Rightarrow(n+2) \times(n+1) \times(n)=60

\Rightarrow(n+2) \times(n+1) \times(n)=5 \times 4 \times 3

\therefore n=3

 

QUESTION 8

If (\mathrm{n}+1) !=90[(\mathrm{n}-1) !], find n

Sol :

(n+1) !=90[(n-1) !]

\Rightarrow(n+1) \times(n) \times(n-1) !=90[(n-1) !]

\Rightarrow(n+1) \times(n)=90

\Rightarrow(n+1) \times(n)=10 \times 9

On comparing, we get:

n=9

 

QUESTION 9

If (n+3) !=56[(n+1) !], find n

Sol :

\Rightarrow(n+3) \times(n+2) \times(n+1) !=56[(n+1) !]

\Rightarrow(n+3) \times(n+2)=56

\Rightarrow(n+3) \times(n+2)=8 \times 7

\Rightarrow n+3=8

\therefore n=5

 

QUESTION 10

If \dfrac{(2 n) !}{3 !(2 n-3) !} and \dfrac{n !}{2 !(n-2) !} are in the ratio 44 : 3, find n

Sol :

\dfrac{(2 n) !}{3 !(2 n-3) !} : \dfrac{n !}{2 !(n-2) !}=44 : 3

\Rightarrow \dfrac{(2 n) !}{3 !(2 n-3) !} \times \dfrac{2 !(n-2) !}{n !}=\dfrac{44}{3}

\Rightarrow \dfrac{(2 n)(2 n-1)(2 n-2)[(2 n-3) !]}{3(2 !)(2 n-3) !} \times \dfrac{2 !(n-2) !}{n(n-1)[(n-2) !]}=\dfrac{44}{3}

\Rightarrow \dfrac{(2 n)(2 n-1)(2 n-2)}{3} \times \dfrac{1}{n(n-1)}=\dfrac{44}{3}

\Rightarrow \dfrac{(2 n)(2 n-1)(2)(n-1)}{3} \times \dfrac{1}{n(n-1)}=\dfrac{44}{3}

\Rightarrow \dfrac{4(2 n-1) n(n-1)}{3} \times \dfrac{1}{n(n-1)}=\dfrac{44}{3}

\Rightarrow \dfrac{4(2 n-1)}{3}=\dfrac{44}{3}

\Rightarrow(2 n-1)=11

\Rightarrow 2 n=12

\Rightarrow n=6

 

QUESTION 11

Prove that:

(i) \dfrac{n !}{(n-r) !}=n(n-1)(n-2) \ldots(n-(r-1))

Sol :

\mathrm{LHS}=\dfrac{n !}{(n-r) !}

=\dfrac{n(n-1)(n-2)(n-3)(n-4) \dots[n-(r-1)][(n-r) !]}{(n-r) !}

=\dfrac{n(n-1)(n-2)(n-3)(n-4) \dots(n-r+1)[(n-r) !]}{(n-r) !}

=n(n-1)(n-2)(n-3)(n-4) \ldots(n-r+1)

=n(n-1)(n-2)(n-3)(n-4) \ldots[n-(r-1)]=\mathrm{RHS}

 

(ii) \dfrac{n !}{(n-r) ! r !}+\dfrac{n !}{(n-r+1) !(r-1) !}=\dfrac{(n+1) !}{r !(n-r+1) !}

Sol :

\mathrm{LHS}=\dfrac{n !}{(n-\mathrm{r}) ! \mathrm{r} !}+\dfrac{n !}{(n-\mathrm{r}+1) !}

=\dfrac{n !}{(n-r) ! r !}+\dfrac{n !}{(n-r+1)[(n-r) !]}

=\dfrac{n !(n-r+1)+n ! r !}{r !(n-r+1)[(n-r) !]}

=\dfrac{n !(n+1)-n ! r !+n ! r !}{r !(n-r+1)(n-r) !}

=\dfrac{n !(n+1)}{r !(n-r+1)(\mathrm{n}-\mathrm{r}) !}

=\dfrac{(n+1 !)}{r !(n-r+1) !}=\mathrm{RHS}

Hence proved 

 

QUESTION 12

Prove that: \dfrac{(2 n+1) !}{n !}=2^{n}[1 \cdot 3 \cdot 5 \ldots(2 n-1)(2 n+1)]

Sol :

\mathrm{LHS}=\dfrac{(2 n+1) !}{n !}

=\dfrac{(2 n+1)(2 n)(2 n-1) \dots .(4)(3)(2)(1)}{n !}

=\dfrac{[(1)(3)(5) \ldots \ldots .(2 n-1)(2 n+1)][(2)(4)(6) \dots \ldots \ldots(2 n)]}{n !}

=\dfrac{2^{n}[(1)(3)(5) \ldots \ldots(2 n-1)(2 n+1)][(1)(2)(3) \ldots \ldots(n)]}{n !}

=\dfrac{2^{n}[(1)(3)(5) \dots \ldots \ldots(2 n-1)(2 n+1)][n !]}{n !}

=2^{n}[(1)(3)(5) \ldots \ldots \ldots(2 n-1)(2 n+1)] =\mathrm{RHS}

Hence proved

Leave a Reply

Your email address will not be published. Required fields are marked *