**Previous Exercise Permutation and Combination – Exercise 4.3 **

#### Permutation and Combination – Exercise 4.4

- Evaluate:

(i) ^{12}P_{4}

(ii) ^{75}P_{2}

(iii) ^{8}P_{8}

(iv) ^{15}P_{1}

(v) ^{38}P_{0}

(i) If ^{n}P_{4} = 20 ^{n}P_{2} find n

(ii)If ^{5}P_{r} = 2. ^{6}P_{r-1} find r

- If
^{n}P_{4 }:^{n}P_{5}= 1:2 find n - If
^{9}P_{5}+ 5.^{ 9}P_{4}=^{10}P_{r} , find r

#### Permutation and Combination – Exercise 4.4 – Solutions

**Evaluate:**

**(i) ^{12}P_{4}**

**(ii) ^{75}P_{2}**

**(iii) ^{8}P_{8}**

**(iv) ^{15}P_{1}**

**(v) ^{38}P_{0}**

Solution:

(i) ^{12}P_{4}

^{n}**P _{r} = ^{n!}/_{(n-r)!}**

= ^{12!}/_{(12 – 4)!}

= ^{12!}/_{8!}

= ^{8! 9x10x11x12}/_{8!}

= 11880

(ii) ^{75}P_{2}

^{n}**P _{r} = ^{n!}/_{(n-r)!}**

= ^{75!}/_{(75-2)!}

= ^{75!}/_{73!}

= ^{73! 74 x 75}/_{73!}

= 5550

(iii) ^{8}P_{8}

^{n}**P _{r} = ^{n!}/_{(n-r)!}**

= ^{8!}/_{(8-8)!}

= ^{8!}/_{0!}

= ^{8!}/_{1}

= 8!

= 40320

(iv) ^{15}P_{1}

^{n}**P _{r} = ^{n!}/_{(n-r)!}**

= ^{15!}/_{(15-1)!}

= ^{14!}/_{14! x 15}

= ^{1}/_{15}

= 0.0667

(v) ^{38}P_{0}

^{n}**P _{r} = ^{n!}/_{(n-r)!}**

= ^{38!}/_{(38 – 0)!}

= ^{38!}/_{38!}

= 1

**2. (i) If ^{n}P_{4} = 20 ^{n}P_{2} find n**

**(ii)If ^{5}P_{r} = 2. ^{6}P_{r-1} find r**

Solution:

(i) ^{n}P_{4} = 20 ^{n}P_{2}

We know, ^{n}P_{r} = ^{n!}/_{(n-r)!}

^{n!}/_{(n-4)!} = 20 x ^{n!}/_{(n-2)!}

^{(n-4)!(n-3)(n-2)(n-1)n}/_{(n-4)!} = 20 x ^{(n-2)! (n – 1)n}/_{(n-2)!}

(n-3)(n-2)(n-1)n = 20 x (n – 1)n

(n -3)(n – 2) = 20

n^{2} – 2n – 3n + 6 = 20

n^{2} – 5n + 6 – 20 = 0

n^{2} – 5n – 14 = 0

n^{2} – 7n + 2n – 14 = 0

n(n – 7) +2(n – 7) = 0

(n + 2)(n – 7) = 0

n = – 2 or n = 7

Therefore, in ^{n}P_{4} = 20 ^{n}P_{2 }, we have n = 7

** **

(ii)If ^{5}P_{r} = 2. ^{6}P_{r-1} , we have to find r

^{5}P_{r} = 2. ^{6}P_{r-1}

^{5!}/_{(5-r)!} = 2 . ^{6!}/_{(6-(r-1))!}

^{5!}/_{(5-r)!} = 2 . ^{6!}/_{(6-r+1)!}

^{5!}/_{(5-r)!} = 2. ^{5! x 6}/_{(7-r)!}

^{1}/_{(5-r)!} = 2. ^{6}/_{(5-r)!(6-r)(7-r)}

1 = 2. ^{6}/_{(6-r)(7-r)}

1 = ^{12}/_{(6-r)(7-r)}

(6 – r)(7 – r) = 12

42 – 6r – 7r + r^{2} = 12

r^{2} – 13r + 30 = 0

r^{2} – 10r – 3r + 30 = 0

r(r – 10) -3(r – 10) = 0

(r – 10) (r – 3) = 0

⇒ r = 10 or r = 3

**If**^{n}P_{4 }:^{n}P_{5}= 1:2 find n

Solution:

^{n}P_{4 }: ^{n}P_{5} = 1:2

^{nP4} / _{nP5} = ^{1}/_{2}

^{n}P_{4}=^{n}P_{5}^{n!}/_{(n-4)!}=^{ n!}/_{(n-5)!}^{n!}/_{(n-5)!(n-4)}=^{n!}/_{(n-5)!}

2 x ^{1}/_{(n – 4)} = 1

2 = (n – 4)

n – 4 = 2

n = 2 + 4 = 6

**If**^{9}P_{5}+ 5.^{ 9}P_{4}=^{10}P_{r} , find r

Solution:

^{9!}/_{(9-5)!} + 5. ^{9!}/_{(9-4)!} = ^{10!}/_{(10 – r)!}

^{9!}/_{4!} + 5. ^{9!}/_{4!x5} = ^{10!}/_{(10 – r)!}

9!(^{1}/_{4!} + ^{1}/_{4!}) = ^{9! x 10}/_{(10 – r)!}

(^{1}/_{4!} + ^{1}/_{4!}) = ^{10}/_{(10 – r)!}

^{1}/_{4!}= 10.^{1}/_{(10 – r)!}

^{1}/_{4!} = 5. ^{1}/_{(10 – r)!}

^{1}/_{5!} = ^{1}/_{(10 – r)!}

5! = (10 – r)!

5 = 10 – r

r = 5

**Next exercise – Permutation and Combination – Exercise 4.5**