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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