Class 11 Maths Ncert Solutions Chapter 7 Ex 7.2 Permutations & Combinations PDF

# Class 11 Maths Ncert Solutions Ex 7.2

## Class 11 Maths Ncert Solutions Chapter 7 Ex 7.2

Q-1: Evaluate:

(i). 8!                                                                    (ii). 4! – 3!

Solution:

(i). 8! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 = 40320

(ii). 4! = 1 × 2 × 3 × 4 = 24

And, 3! = 1 × 2 × 3 = 6

Therefore, 4! – 3! = 24 – 6 = 18

Q-2: Is 3! + 4! = 7! ?

Solution:

Now, 3! = 1 × 2 × 3 = 6

4! = 1 × 2 × 3 × 4 = 24

And, 7! = 1 × 2 × 3 × 4 × 5 × 6 × 7 = 5040

Therefore, 4! – 3! ≠ 7!

Q-3: Compute:

9!5!×3!$\frac{ 9! }{ 5!\; \times \;3! }$

Solution:

9!5!×3!$\frac{ 9! }{ 5!\; \times \;3! }$ = 1×2×3×4×5×6×7×8×91×2×3×4×5×1×2×3$\frac{ 1 \;\times \;2 \;\times \;3 \;\times \;4\; \times \;5\; \times \;6\; \times \;7\; \times \;8\; \times \;9 }{ 1\; \times \;2\; \times \;3\; \times \;4\; \times \;5\; \times \;1\; \times \;2\; \times \;3 }\\$

Therefore, 9!5!×3!$\frac{ 9! }{ 5! \;\times \;3! }$ = 392

Q-4. If 15!+16!=a7!$\frac{ 1 }{ 5! } + \frac{ 1 }{ 6! } = \frac{ a }{ 7! }$, find the value of a.

Solution:

15!+16!=a7!$\;\frac{ 1 }{ 5! } + \frac{ 1 }{ 6! } = \frac{ a }{ 7! }\\$

$\\\Rightarrow$   15!+16×5!=a7×6×5!$\frac{ 1 }{ 5! } + \frac{ 1 }{ 6 \;\times \;5! } = \frac{ a }{ 7 \;\times \;6 \times \;5! }\\$

$\\\Rightarrow$   15!+[1+16]=a7×6×5!$\frac{ 1 }{ 5! } + \left [ 1 + \frac{ 1 }{ 6 } \right ] = \frac{ a }{ 7 \;\times \;6 \times \;5! }\\$

1+16=a7×6$\\\Rightarrow 1 + \frac{ 1 }{ 6 } = \frac{ a}{ 7 \;\times \;6 }\\$

$\\\Rightarrow$   76=a7×6$\frac{ 7 }{ 6 } = \frac{ a }{ 7 \times 6 }\\$

a=7×7×66$\Rightarrow a = \frac{ 7 \;\times 7 \;\times 6 }{ 6 }\\$

Therefore, a = 49

Q-5: Evaluate: a!(ar)!$\frac{ a! }{ \left ( a – r \right )! }$, when

(i). a = 5, r = 1                                                       (ii). a = 8, r = 4

Solution:

(i)  a = 5, r = 1:         [ Given ]

a!(ar)!$\\\frac{ a! }{ \left ( a\; – \;r \right )! }$ = 5!(51)!$\frac{ 5! }{ \left ( 5 \;-\; 1 \right )! }$= 5!4!$\frac{ 5! }{ 4! }$= 5×4!4!$\frac{ 5\; \times \;4! }{ 4! }\\$

Therefore, a!(ar)!$\frac{ a! }{ \left ( a \;-\; r \right )! }\;$ = 5

(ii) a!(ar)!$\;\frac{ a! }{ \left ( a \;-\; r \right )! }$ = 8!(84)!$\frac{ 8! }{ \left ( 8 \;-\; 4 \right )! }$= 8!4!$\frac{ 8! }{ 4! }\\$

= 8×7×6×5×4!4!$\\\frac{ 8\; \times \;7 \times 6 \;\times \;5 \times \;4! }{ 4! }\\$= 8 × 7 × 6 × 5 = 15120

Therefore, a!(ar)!$\\\frac{ a! }{ \left ( a \;-\; r \right )! }$ = 15120