(a) Number of permutations of n things, taken r at a time, when a particular thing is to be always included in each arrangement
= r n-1 Pr-1
(b) Number of permutations of n things, taken r at a time, when a particular thing is fixed: = n-1 Pr-1
(c) Number of permutations of n things, taken r at a time, when a particular thing is never taken: = n-1 Pr.
(d) Number of permutations of n things, taken r at a time, when m specified things always come together = m! x ( n-m+1) !
(e) Number of permutations of n things, taken all at a time, when m specified things always come together = n ! - [ m! x (n-m+1)! ]
Example: How many words can be formed with the letters of the word OMEGA when:
(i) O and A occupying end places.
(ii) E being always in the middle
(iii) Vowels occupying odd-places
(iv) Vowels being never together.
Ans.
(i) When O and A occupying end-places
=> M.E.G. (OA)
Here (OA) are fixed, hence M, E, G can be arranged in 3! ways
But (O,A) can be arranged themselves is 2! ways.
=> Total number of words = 3! x 2! = 12 ways.
(ii) When E is fixed in the middle
=> O.M.(E), G.A.
Hence four-letter O.M.G.A. can be arranged in 4! i.e 24 ways.
(iii) Three vowels (O,E,A,) can be arranged in the odd-places (1st, 3rd and 5th) = 3! ways.
And two consonants (M,G,) can be arranged in the even-place (2nd, 4th) = 2 ! ways
=> Total number of ways= 3! x 2! = 12 ways.
(iv) Total number of words = 5! = 120!
If all the vowels come together, then we have: (O.E.A.), M,G
These can be arranged in 3! ways.
But (O,E.A.) can be arranged themselves in 3! ways.
=> Number of ways, when vowels come-together = 3! x 3!
= 36 ways
=> Number of ways, when vowels being never-together
= 120-36 = 84 ways.
Number of Combination of n different things, taken r at a time is given by:-
nCr= n! / r ! x (n-r)!
Proof: Each combination consists of r different things, which can be arranged among themselves in r! ways.
=> For one combination of r different things, number of arrangements = r!
For nCr combination number of arrangements: r nCr
=> Total number of permutations = r! nCr ---------------(1)
But number of permutation of n different things, taken r at a time
= nPr -------(2)
From (1) and (2) :
nPr = r! . nCr
or n!/(n-r)! = r! . nCr
or nCr = n!/r!x(n-r)!
Note: nCr = nCn-r
or nCr = n!/r!x(n-r)! and nCn-r = n!/(n-r)!x(n-(n-r))!
= n!/(n-r)!xr!
Restricted Combinations
(a) Number of combinations of n different things taken r at a time, when p particular things are always included = n-pCr-p.
(b) Number of combination of n different things, taken r at a time, when p particular things are always to be excluded = n-pCr
Example: In how many ways can a cricket-eleven be chosen out of 15 players? if
(i) A particular player is always chosen,
(ii) A particular is never chosen.
Ans:
(i) A particular player is always chosen, it means that 10 players are selected out of the remaining 14 players.
=. Required number of ways = 14C10 = 14C4
= 14!/4!x19! = 1365
(ii) A particular players is never chosen, it means that 11 players are selected out of 14 players.
=> Required number of ways = 14C11
= 14!/11!x3! = 364
(iii) Number of ways of selecting zero or more things from n different things is given by:- 2n-1
Proof: Number of ways of selecting one thing, out of n-things = nC1
Number of selecting two things, out of n-things =nC2
Number of ways of selecting three things, out of n-things =nC3
Number of ways of selecting n things out of n things = nCn
=>Total number of ways of selecting one or more things out of n different things
= nC1 + nC2 + nC3 + ------------- + nCn
= (nC0 + nC1 + -----------------nCn) - nC0
= 2n 1 [ nC0=1]
Example: John has 8 friends. In how many ways can he invite one or more of them to dinner?
Ans. John can select one or more than one of his 8 friends.
=> Required number of ways = 28 1= 255.
(iv) Number of ways of selecting zero or more things from n identical things is given by :- n+1
Example: In how many ways, can zero or more letters be selected form the letters AAAAA?
Ans. Number of ways of :
Selecting zero 'A's = 1
Selecting one 'A's = 1
Selecting two 'A's = 1
Selecting three 'A's = 1
Selecting four 'A's = 1
Selecting five 'A's = 1
=> Required number of ways = 6 [5+1]
(V) Number of ways of selecting one or more things from p identical things of one type q identical things of another type, r identical things of the third type and n different things is given by :-
(p+1) (q+1) (r+1)2n 1
Example: Find the number of different choices that can be made from 3 apples, 4 bananas and 5 mangoes, if at least one fruit is to be chosen.
Ans:
Number of ways of selecting apples = (3+1) = 4 ways.
Number of ways of selecting bananas = (4+1) = 5 ways.
Number of ways of selecting mangoes = (5+1) = 6 ways.
Total number of ways of selecting fruits = 4 x 5 x 6
But this includes, when no fruits i.e. zero fruits is selected
=> Number of ways of selecting at least one fruit = (4x5x6) -1 = 119
Note :- There was no fruit of a different type, hence here n=o
=> 2n = 20=1
(VI) Number of ways of selecting r things from n identical things is 1.
Example: In how many ways 5 balls can be selected from 12 identical red balls?
Ans. The balls are identical, total number of ways of selecting 5 balls = 1.
Example: How many numbers of four digits can be formed with digits 1, 2, 3, 4 and 5?
Ans. Here n = 5 [Number of digits]
And r = 4 [ Number of places to be filled-up]
Required number is 5P4 = 5!/1! = 5 x 4 x 3 x 2 x 1
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