11th NCERT Permutations And Combinations
Hi

Exercise 7.2 Questions 5

Question (1)

Evaluate (i) 8! (ii) 4! – 3!

Solution

(i) 8! = 8 × 7 × 6 ×5 × 4 × 3 × 2× 1
= 40320.
(ii) 4! - 3!
= 4×3×2×1 - 3×2× 1
= 24 - 6
= 18

Question (2)

Is 3! + 4! = 7!?

Solution

LHS = 3! + 4!
= 3×2 + 4× 3 × 2
= 6 + 24
= 30
RHS = 7!
= 7×6×5×4×3×2×1
= 5040
≠ LHS.
So 3! + 4! ≠ 7!.

Question (3)

Compute \[\frac{{8!}}{{6!\; \times \,2!}}\]

Solution

\[\frac{{8!}}{{6!\; \times \,2!}}\] \[ = \frac{{8 \times 7 \times 6!}}{{6!\, \times 2}}\] \[ = 28\]

Question (4)

If ,\[\frac{1}{{6!}} + \frac{1}{{7!}} = \frac{x}{{8!}}\] find x.

Solution

\[\frac{1}{{6!}} + \frac{1}{{7!}} = \frac{x}{{8!}}\] \[\frac{1}{{6!}} + \frac{1}{{7\, \times 6!}} = \frac{x}{{8 \times 7 \times 6!}}\] \[1 + \frac{1}{7} = \frac{x}{{56}}\] \[\frac{8}{7} = \frac{x}{{56}}\] \[x = 64\]

Question (5)

Evaluate,\[\frac{{n!}}{{\left( {n - r} \right)!}}\] when (i) n = 6, r = 2 (ii) n = 9, r = 5

Solution

(i) n = 6 , r = 2
\[\frac{{n!}}{{\left( {n - r} \right)!}}\] \[ = \frac{{6!}}{{\left( {6 - 2} \right)!}}\] \[ = \frac{{6!}}{{4!}}\] \[ = \frac{{6 \times 5 \times 4!}}{{4!}}\] \[ = 30\] (ii) n = 9, r = 5
\[\frac{{n!}}{{\left( {n - r} \right)!}}\] \[ = \frac{{9!}}{{\left( {9 - 5} \right)!}}\] \[ = \frac{{9!}}{{4!}}\] \[ = \frac{{9 \times 8 \times 7\; \times 6 \times 5 \times 4!}}{{4!}}\] \[ = 15120\]
7.1
Exercise 7.3