Contents
Exercise 13.2
Question 1
निम्नलिखित का परिकलन कीजिए
[Evaluate the following :]
(i) 7!
Sol :
=7×6×5×4×3×2×1
=5040
(ii) 5!
Sol :
(iii) 8!
Sol :
(iv) 8!-5!
Sol :
=8×7×6×5×4×3×2×1-5×4×3×2×1
=40320-120
=40200
(v)4!-3!
Sol :
(vi) 7!-5!
Sol :
(vii) $\frac{6!}{5!}$
Sol :
$=\frac{6 \times 5 !}{5 !}$
=6
(viii) $\frac{7 !}{5 !}$
Sol :
(ix) $\frac{81}{6 ! 2 !}$
Sol :
$=\frac{8 \times 7 \times 6 !}{6 ! \times 2\times 1}$
=28
(x) $\frac{9 !}{4 ! 5 !}$
Sol :
(xi) $\frac{12 !}{(10 !) 2 \times 1}$
Sol :
$=\frac{12 \times 11 \times 10 !}{10! \times 2 \times 1}$
=66
Question 2
परिकलन कीजिए
[Compute]
(i) (3!)(5!)
Sol :
=3×2×1×5×4×3×2×1
=720
(ii) $\frac{20 !}{18 !(20-18) !}$
Sol :
$=\frac{20 \times 19 \times 18 !}{18 !~ 2 \times 1}$
=190
(iii) $\frac{1}{5 !}+\frac{1}{6 !}+\frac{1}{7 !}$
Sol :
$=\frac{42+7+1}{7 !}$
$=\frac{50}{5040}=\frac{5}{504}$
Question 3
निकालिए (Evaluate )$\frac{n !}{r !(n-r) !}$ जब(when)]
(i) n=7 , r =3
Sol :
$=\frac{7 !}{3 !(7-3) !}$
$=\frac{7 !}{3 !~ 4 !}$
$=\frac{7 \times 6 \times 5 \times 4 !}{3 \times 2 \times 1 \times 4!}$
=35
(ii) n=15 , r =12
Sol :
(iii) n=5 , r=2
Sol :
Question 4
निकालिए [Evaluate $\frac{n !}{(n-r) !}$, जब (when)
(i) n=9, r=5
(ii) n=6, r=2
Sol :
Question 5
निम्नलिखित को क्रमगुणित में बदलिए
[Convert the following into factorials]
(i) 1.3.5.7.9.11
Sol :
$=\frac{1.2. 3.4 .5 .6 .7 .8 . 9.10 .11}{2.4 .6 .8 .10}$
$=\frac{11 !}{2^{5} \cdot(1.2.3.4.5)}$
$=\frac{11 !}{2^{5} \cdot 5 !}$
(ii) (n+1)(n+2)(n+3)…2n
Sol :
$=\frac{1.2 .3 \cdot \ldots n(n+1)(n+2)(n+3) \ldots 2 n}{1.2.3 \ldots \ldots n}$
$=\frac{(2 n) !}{n !}$
Question 6
सत्य और असत्य बताइए
[State whether ‘true’ or ‘false’]
(i) 2!+3!=5!
Sol :
L.H.S
=2!+3!
=2×1+3×2×1
=2+6
=8
False
(ii) 2!×3!=6!
Sol :
L.H.S
=2!×3!
=2×1×3×2×1
=12
False
(iii) $\frac{8 !}{4 !}=2 !$
Sol :
(iv) 5!-3!=2!
Sol :
(v) 3!+4!=7!
Sol :
Question 7
x ज्ञात कीजिए यदि
[Find x if :]
(i) $\frac{1}{8 !}+\frac{1}{9 !}=\frac{x}{10 !}$
Sol :
$\frac{9+1}{9 !}=\frac{x}{10 !}$
$\frac{10}{9!}=\frac{x}{10 \times 9 !}$
100=x
(ii) $\frac{1}{6 !}+\frac{1}{7 !}=\frac{x}{8 !}$
Sol :
Question 8
n का मान निकालिए यदि
[Find the value of n if]
(i) (n+1)!=12.(n-1)!
Sol :
$(n+1) \cdot n \cdot(n – 1) !=12 \cdot(n-1) !$
(n+1).n=12
(n+1)n=4×12
n+1=4
n=3
and n=3
(ii) 2n!n!=(n+1)(n-1)!(2n-1)!
Sol :
$=(n+1)(n-1) !(2 n-1) !$
$2 n \cdot(2 n-1) ! n \cdot(n-1) !=(n+1)(n-1) !(2 n-1) !$
$2 n^{2}=n+1$
$2 n^{2}-n-1=0$
$2 n^{2}-2 n+n-1=0$
$2 n(n-1)+1(n-1)=0$
$(n-1)(2 n+1)=0$
$\begin{array}{r|l}n-1=0 & 2 n+1=0 \\ n=1 & 2 n=-1\\&n=\frac{-1}{2}\end{array}$
Question 9
यदि [If] $\frac{n !}{2 !(n-2) !}$ और (and) $\frac{n !}{4 !(n-4) !} 2: 1$ में है, n का मान ज्ञात कीजिए (are in ratio 2 : 1, find the value of n)
Sol :
$\frac{n !}{2 !(n-2) !}: \frac{n !}{4 !(n-4)!}=2:1$
$\frac{\frac{n!}{2 !(n-2) !}}{\frac{n!}{4 !(n-4) !}}=\frac{2}{1}$
$\frac{4 !(n-4) !}{2 !(n-2) !}=\frac{2}{1}$
$\frac{4 \times 3 \times 2 !(n-4) !}{2 ! \times(n-2)(n-3)(n-4) !}=\frac{2}{1}$
$\frac{12}{(n-2)(n-3)}=\frac{2}{1}$
2(n-2)(n-3)=12
(n-2)(n-3)=6
(n-2)(n-3)=3×2
$\begin{array}{r|l}n-2=3& n-3=2 \\ n=5 & 2 n=5\end{array}$
Question 10
दिखाइए कि (Show that) n! (n+2)=n!+(n+1)!
Sol :
L.H.S
=n!(n+2)
=n![(n+1)+1]
=n!(n+1)+n!
=(n+1)!+n!
Question 11
x का मान निकालिए यदि
[Find the value of x if]
$\frac{(x+2) !}{(2 x-1) !} \cdot \frac{(2 x+1) !}{(x+3) !}=\frac{72}{7}$
जहाँ (where x?N)
Sol :
$\frac{(x+2) !}{(2 x-1) !} \cdot \frac{(2 x+1) !}{(x+3) !}=\frac{72}{7}$
$\frac{(x+2) !}{(2 x-1) !} \times \frac{(2 x+1) \cdot 2 x(2 x-1) !}{(x+3)(x+2) !}=\frac{72}{7}$
$14 x(2 x+1)=72(x+3)$
$28 x^{2}+14 x=72 x+216$
$28 x^{2}+14 x-72 x-216=0$
$28 x^{2}-58 x-216=0$
$2\left(14 x^{2}-29 x-108\right)=0$
$14 x^{2}-29 x-108=0$
$14 x^{2}-56 x+27 x-108 =0$
$14 x(x-4)+27(x-4) =0$
$(x-4)(14 x+27) =0$
$\begin{array}{r|l}x-4=0 & 14 x+27=0 \\ x=4 &14 x=-27 \\ &x=\frac{-27}{14}\end{array}$
∵ x?N
∴ x=4