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DIFFERENT FORMULAS ON PERMUTATION:
Theorem 1:
To find the number of permutations of n different things taken r at a time (r ≤ n).
The number of permutations of n different things, taken r at a time, none
of things being repeated in any permutation. Suppose, we have r blank
spaces and n ( ≥ r) different things. Then the required number of
permutations will be obviously the same as the number of ways in which r
blank places can be filled up with n different things. Clearly, the first
place can be filled up in n different ways, for any one of the n different
things can be placed there. So when the first place has been filled up in
any one of these n ways, the second place can be filled up in (n-1) ways,
for any one of the remaining (n - 1) things can be taken to fill up the
second place. Clearly, each way of filling up the first place can be
associated with each of the (n - 1) ways of filling up the second place.
Therefore, the first two places can be filled up in n (n - 1) ways.
Again, when the first two places have been filled up in any one of these
n (n - 1) ways, the third place can be filled in (n - 2) ways, for any one
of the remaining (n - 2) things can be placed there. As each way of filling
up the first two places can be associated with each of the (n - 2) ways of
filling up the third place, the first, second and third places can be filled
up in n (n - 1) (n - 2) ways.
Continuing similarly and noting that a new factor is introduced with each of
the new place filled up and the number of factors at any stage is equal to
the number of places filled up, we can say that the number of ways in which
r blank places can be filled up by n different things is
n (n - 1) (n - 2) ..... to r factors
where rth factor = n - (r - 1) = n - r + 1 [Since the second factor = n - 1 = n - (2 - 1); the third factor = n - 2 = n - (3 - 1) and so on].
Therefore, the required number of permutations of n different things taken r at a time
= n (n - 1) (n - 2) ........ (n - r + 1). ..... (1)
Corollary: The number of permutations of n different things taken all at a time (or permutations of n different things among themselves)
= n (n - 1) (n - 2) ........ (n - n + 1) [Using r = n in (1)]
= n (n - 1) (n - 2) ........3 ⋅ 2 ⋅ 1 ......(2)
Notations:
Theorem 2:
To find the number of permutations of n things taken them all at a time, when
p of the things are alike of one kind, q of the things are alike of another
kind, r of the things are alike of a third kind and the rest are all different.
We denote the n things by n letters and assume that p of them to be a, q of
them to be b, r of them to be c and the rest to be different.
Let consider x to denote the required number of permutations.
Obviously, there are p a's in each of the x permutations. Now, let us consider
any one of these x permutations and replace the p a's by p new letters a
1, a2 ......ap different from each other and also
different from each other and also different from the remaining (n - p) letters.
Clearly these new p letters can be arranged among themselves in p! ways. Therefore,
if this change is made in each of the x permutations, we will get x * p! permutations
in which p a's are will be all different.
Let us consider any one of these x * p! permutations and replace the q b's by a new
letters b1, b2 ....., bq different from each other and also different from the rest.
Obviously, these q new letters can be arranged among themselves in q! ways. Therefore,
making such change in each of the x * p! permutations we will get x * p! * q! permutations
in which p a's and q b's will be all different.
Therefore, replacing the r c's by r new letters c1,c2,......,cr
and then per mutating them
we will get x * p! * q! * r! permutations.
So, clearly the n letters are now all different. Therefore, x * p * q! * r! = the number of permutations
of n different things taken all at a time = n!
Theorem 3:
To find the number of permutations of n different things taken r at a time, when each
thing may be repeated once, twice, ...., upto r times in any arrangement.
Supposing that we have r blank places and n different things where each thing
may be taken once, twice, ..., upto r times. Then the required number of permutations
is the same as the number of ways in which r blank places can be filled up by n
different things
Then, the first place can be filled up in n ways, because any of the n may be
put in it. When the first place is filled up, the second place can also be
filled up in n ways, since the same letters may be used again. Thus the first
two places can be filled up in n * n ways, i.e. n2 ways. The third place can
also be filled up in n ways and, therefore, the first three places can be
filled up in n2 * n = n3 ways.
Proceeding in this way and noticing that at any stage the index of n is always
the same as the number of places filled up. Therefore, r blank places can be
filled up by n different things in nr ways. Therefore, the required number
of ways in which the r places can be filled is nr.
Example 1:
If 4 - x P 2 = 6, find the value of x?
Solution:
Since n P 2 = n (n - 1) hence, 4 - x P 2 = 6 gives, (4 - x) (4 - x - 1) = 6
Or, y(y - 1) = 6 [where 4 - x = y]
Or, y2 - y - 6 = 0
Or, (y - 3) (y + 2) = 06
Therefore, either, y - 3 = 0 i.e., y = 3 Or, y + 2 = 0 i.e., y = -2
Since y = 4 - x is positive, hence we take y = 3
Or, 4 - x = 3 [Since, y = 4 - x] or, x = 1
Since n P 2 = n (n - 1) hence, 4 - x P 2 = 6 gives, (4 - x) (4 - x - 1) = 6
Or, y(y - 1) = 6 [where 4 - x = y]
Or, y2 - y - 6 = 0
Or, (y - 3) (y + 2) = 06
Therefore, either, y - 3 = 0 i.e., y = 3 Or, y + 2 = 0 i.e., y = -2
Since y = 4 - x is positive, hence we take y = 3
Or, 4 - x = 3 [Since, y = 4 - x] or, x = 1
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