**Example 1 :**

If ^{(n - 1)} P_{3} : ^{n} P_{ 4 } = 1 : 10, find the value of n.

**Solution :**

^{n}P_{r} = n!/(n-r)!

[(n-1)!/(n-4)!] / [n!/(n-4)!]_{ } = (1/10)

[(n-1)!/(n-4)!] ⋅ [(n-4)!/n!]_{ } = (1/10)

[(n-1)!/n(n-1)!]_{ } = (1/10)

1/n = 1/10

n = 10

Hence the value of n is 10.

**Example 2 :**

If ^{10}P_{r−1} = 2 ⋅ ^{6}P_{r}, find r.

**Solution :**

10!/(11-r)! = 2 ⋅ (6!/(6-r)!)

10⋅9⋅8⋅7⋅6!/(11-r)(10-r)(9-r)(8-r)(7-r)(6-r)! = 2⋅(6!/(6-r)!)

10⋅9⋅8⋅7/(11-r)(10-r)(9-r)(8-r)(7-r) = 2

(11-r)(10-r)(9-r)(8-r)(7-r) = 10⋅9⋅4⋅7

(11-r)(10-r)(9-r)(8-r)(7-r) = 5 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 4 ⋅ 7

(11-r)(10-r)(9-r)(8-r)(7-r) = 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3

11 - r = 7

11 - 7 = r

r = 4

Hence the value of r is 4.

**Example 3 :**

(i) Suppose 8 people enter an event in a swimming meet. In how many ways could the gold, silver and bronze prizes be awarded?

(ii) Three men have 4 coats, 5 waist coats and 6 caps. In how many ways can they wear them?

**Solution :**

(i) 8 people have equal chances to get 1st prize gold, 7 people are having equal chances of getting the 2nd prize silver, 6 people are having equal chances of getting 3rd prize.

Hence the total number of ways = 8 ⋅ 7 ⋅ 6

= 336 ways

(ii) Three men have 4 coats, 5 waist coats and 6 caps. In how many ways can they wear them?

**Solution :**

1^{st} man can wear any of the 4 coats.

2^{nd} man can wear any of the remaining 3 coats.

3^{rd} man can wear any of the remaining 2 coats.

So, number of ways in which 3 men can wear 4 coats

= 4 ⋅ 3⋅ 2

= 24

Similarly,

Number of ways in which 3 men can wear 5 waistcoats

= 5 ⋅ 4 ⋅ 3

= 60

Number of ways in which 3 men can wear 6 caps

= 6 ⋅ 5 ⋅ 4

= 120

Hence, required number of ways

= 24 ⋅ 60 ⋅ 120

= 172800

Hence the total number ways is 172800.

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