Page 15 - Life Insurance Today March 2016
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which p objects are similar. Here the p similar objects selected out of 5 candidates. In how many ways can
may be denoted by a1, a2, a3 …..ap and the remaining the selection be made?
objects by b, c, d, e…..In this arrangement a1, a2, a3
…..ap objects can be arranged without disturbing b, Here, we may find that the order in which we select
c, d, …. in p! ways. Since a1, a2, a3 …..ap are similar, the two candidates is immaterial. Therefore, the
these p! different arrangements become always number of ways of selection
identical. Therefore, the number of arrangements with
p like objects and other unlike objects is n! / p! 5X4
= = 10
Similarly, if out of n objects, p are of one kind, q are
of other kind, r are of third kind and so on, then the 2
number of ways in which n objects can be arranged If a, b, c, d & e are the candidates; the selection of
in a row is given by - two candidates can be either - ab, ac, ad, ae, bc, bd,
be, cd, ce, de. Here, combinations ab and ba are
n! treated as a single selection as the two candidates are
= a definite number. same. Similarly for ac and ca, ae and ea, bc and cb,
etc. are all equal selection.
p!q!r!..
The number of selections or combinations of n (unlike)
Example: Find the number of distinct permutations of objects taken r at a time is r! nCr, and is given by -
the letters of the word "COMMITTEE". Find also the
number of arrangements in which - nCr = n (n-1) (n-2) …. (n-r+1) = n!
(i) The two M's come together ;
1X2X3….Xr r! (n - r) !
For each of nCr selections of r objects, r objects can
(ii) The two M's do not come together; be arranged in r! ways. Thus total number of
(iii) The two M's and the two T's come together. arrangements of n (unlike) objects taken r at a time is
Solution: The word "COMMITTEE" contains 9 letters r! nCr - but the no. of such arrangements is given by
of which M, T and E are repeated twice. Therefore,
the number of distinct arrangements nPr
9! n!
= = 45,360 = = r! nCr
2! 2! 2! (n - r)!
(i) In the first case, we need to treat two M's coming
Therefore, when r = n, nCr = nCn = n! =1
together as a single letter. Therefore, permutation
s of 8 letters with T and E repeating twice will be 0! n!
given by
So selection of all n objects out of n objects can be
8! 40320
= 2! 2! = 4 = 10080. done only in one way and when r=0, we have, nC0
n!
= n!0!
(ii) In the second case the number of arrangements So, nC0 = 1, for r=0 i.e. when r objects are selected
in which two M's do not come together = 45,360 from n (unlike) objects, (n-r) objects are left out.
- 10,080 = 35,280. Therefore, number of selections of r objects from n,
viz. nCr is the same as the number of selections of (n-
(iii) In the third case we need to treat two M coming r) from n, i.e. nCr = nCn-r.
together as a single letter so also for the two T F. Mortality concept based on theory of
probability used in life insurance
coming together. Therefore, here it is the premium rating:
permutation of 7 letters with E repeating twice Mortality experience of the general population which is
commonly known as census mortality is not directly of
may be given by much use to an insurer whose main interest is always to
7! 5040
== = 2520.
2! 2
(7) Theory of Combinations: Say two candidates are to be
"To be yourself in a world that is constantly trying to make you something else is the greatest accomplishment."
Life Insurance Today March 2016 15
