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

• Last Updated : 17 Dec, 2020

### i) 8!

Solution:

8! = 8 Ã— 7 Ã— 6 Ã— 5 Ã— 4 Ã— 3 Ã— 2 Ã— 1

= 56 Ã— 30 Ã— 12 Ã— 2

= 1680 Ã— 24

= 40320

### ii) 4! â€“ 3!

Solution:

4! -3! = 4 Ã— 3! – 3!

= 3! Ã— (4 – 1)

= 3! Ã— 3

= 6 Ã— 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 Ã— 6 Ã— 5 Ã— 4 Ã— 3 Ã— 2 Ã— 1

= 42 Ã— 20 Ã— 6

= 42 Ã— 120

= 5040

As l.H.S â‰  R.H.S

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

### Question 3. Compute 8!/(6! Ã— 2!)

Solution:

8!/(6! Ã— 2!)

= 8 Ã— 7 Ã— 6 ! / (6 ! Ã— 2!)

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

= 8 Ã— 7/2 Ã— 1

= 4 Ã— 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 Ã— 8!/7 ! = y

8 Ã— 8 Ã— 7! / 7! = y

8 Ã— 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 Ã— 5 Ã— 4!/4!

= 6 Ã— 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 Ã— 8 Ã— 7 Ã— 6 Ã— 5 Ã— 4!/4!

= 9 Ã— 8 Ã—7 Ã— 6 Ã— 5

= 72 Ã— 7 Ã— 30

= 72 Ã— 210

= 15120

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