# Class 11 NCERT Solutions – Chapter 7 Permutations And Combinations – Exercise 7.2

• Last Updated : 17 Dec, 2020

### i) 8!

Solution:

8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

= 56 × 30 × 12 × 2

= 1680 × 24

= 40320

### ii) 4! – 3!

Solution:

4! -3! = 4 × 3! – 3!

= 3! × (4 – 1)

= 3! × 3

= 6 × 3

= 18

### Question 2. Is 3! + 4! = 7! ?

Solution:

L.H.S = 3! + 4!

= 3! + 4 * 3!

= 3! * (1 + 4)

= 3! * 5

= 60

R.H.S = 7!

= 7 × 6 × 5 × 4 × 3 × 2 × 1

= 42 × 20 × 6

= 42 × 120

= 5040

As l.H.S ≠ R.H.S

NO, 3!+ 4! is not equal to 7!

### Question 3. Compute 8!/(6! × 2!)

Solution:

8!/(6! × 2!)

= 8 × 7 × 6 ! / (6 ! × 2!)

Now, both 6! on numerator and denominator will be cancelled

= 8 × 7/2 × 1

= 4 × 7

= 28

### Question 4. If (1/6!) + (1/7!) = (y/8!), Find y.

Solution:

L.C.M of 6! and 7! is 7!

We can write the equation like this,

(7 + 1)/7! = (y/8!)

8 × 8!/7 ! = y

8 × 8 × 7! / 7! = y

8 × 8 = y

y = 64

Hence, value of y is 64

### i) n = 6, r = 2

Solution:

n!/(n – r)!

Here n = 6 and r = 2, we have

6!/(6 – 2)!

= 6!/4!

= 6 × 5 × 4!/4!

= 6 × 5

= 30

### ii) n = 9, r = 5

Solution:

n!/(n – r)!

Here n = 9 and r = 5, we have

9!/(9 – 5)!

= 9!/4!

= 9 × 8 × 7 × 6 × 5 × 4!/4!

= 9 × 8 ×7 × 6 × 5

= 72 × 7 × 30

= 72 × 210

= 15120

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