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Derangements and Multinomial Theorem
Any change in the existing order of things is called a derangement.
If 'n' things are arranged in a row, the number of ways in which they can, be deranged so that none of them occupies its original place is
and it is denoted by D(n).
A question on derangement can be of the following kind:
Supposing 4 letters are placed in 4 different envelopes. In how many ways can be they be taken out from their original envelopes and distributed among the 4 different envelopes so that no letter remains in its original envelope?
Using the formula for the number of derangements that are possible out of 4 letters in 4 envelopes, we get the number of ways as :
4!(1 - 1 + 1/2! - 1/3! + 1/4!) = 24(1 - 1 + 1/2 - 1/6 + 1/24) = 9.
Let x1, x2, ......, xm be integers. Then number of solutions to the equation
x1 + x2 + ... + xm = n ... (1)
subject to the conditions a1 < x1 < b1, a2 < x2 < b2, ..., am < xm < bm ... (2)
is equal to the coefficient of xn in
This is because the number of ways in which sum of m integers in (1) subject to given conditions (2) equals n is the same as the number of times xn comes in (3). Using this we get the number of non negative integral solutions of (1) is given by n+m-1Cm-1 and number of positive integral solutions of (1) is given by n-1Cm-1.
In how many different ways three persons A, B, C having 6, 7 and 8 one rupee coins respectively can donate Rs.10 collectively.
The number of ways in which they can denote Rs.10 is the same as the number of solutions of the equation
x1 + x2 + x3 = 10
subject to conditions 0 < x1 < 6, 0 < x2 < 7, 0 < x3 < 8
Hence the required number of ways
= coefficient of x10 in (1+x+x2+...+x6)(1+x+x2+...+x7)(1+x+x2+...+x8)
= coefficient of x10 in (1-x7)(1-x8)(1-x9)(1-x)-3
= coefficient of x10 in (1-x7-x8-x9)(1+3C1x + 4C2x2+ 5C3x3+...+12C10x10)
(ignoring powers higher than 10)
= 12C2 - 5C3 - 4C2 - 3C1
= 66 - 10 - 6 - 3 = 47.
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