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Pascal's Triangle.(
Each number in the triangle is the sum of the two directly above it)
Source: Wikipedia.org |
(a + b)2 = a2 + 2ab + b2
The above expression is something that everyone is familiar
with. However, does it only signify a convenient method of computing the whole
square of a number?
Let us bring in the context of events and probability and
see what we learn.
Suppose we have an event with 2 possible results: ‘a’ &
‘b’. (eg, if the event is a toss of a coin, then ‘a’ could be the result of
having obtained a heads, and ‘b’, a tails).
If we toss the coin once, the outcome is 'a' OR 'b' :- (a + b).
If we toss the coin twice, then the result is: (a+b)(a+b)
The above mathematical expression efficiently represents the different permutation
and combination of all outcomes possible.
If we revisit the equation for (a+b)2, we’d
see that in the RHS, each term represents different sets of combination possible, where the coefficient of each term represent the number of
permutations of each combination.
So, the end result in this case is either
both heads (a.a) OR 2 combination of heads and tails occurring in different
order (a.b + b.a) OR both tails (b.b).
Similarly, if we toss the coin thrice:
(a + b)3 = a3 + 3a2b + 3ab2
+ b3
Again, each term above on RHS represents each possible combination of results, with the additional info that it occurs 'its coefficient' no. of times
Finding the coefficients of each term is easy using Binomial
expansion.
(a + b)n = C(n,0)an + C(n,1)an-1b
+ C(n,2)an-2b2 + .... + C(n,n)bn
The little piece of knowledge in the above article is enough to give us a good perspective to solve
the below example problems:
Example 1: If a fair coin is tossed 10 times, what is the
probability of obtaining 2 heads and 8 tails.
All we are looking for is the
coefficient of a2b8 in the expansion of (a+b)10 => C(10,2) or C(10,8) = 45.
Total sample space = 210
Prob = 45/210
Example 2: Find out the number of terms in the expansion
of (A + B + C)20
We know that each term (result set of outcomes) is
a different combination of total no. of A,B and Cs.
In each of the terms, the
total no. of all of A, B and Cs = 20.
ie, (Total As) + (Total Bs) + (Total Cs)
= 20.
So the question boils down to different non-negative integral solns for:
a + b + c = 20
Above equation has C(22,2) solns = 231
(Refer previous article on P&C Concept 1 for learning how to deduce this result)
Example 3: What is the sum of coefficients of all the terms in
the expansion
of (a + b + c)20
In other words, the question asks for total no. of all possible results (Regardless of the permutation orders) out of an event of 3 outcomes repeated 20 times.
So the answer is nothing but: (No. of different outcomes)20 = 320
To explain more, (a+b+c)20 = k1a20 + k2a19b
+ k3a18bc + .... + k231c20
Where ki are different coefficients.
Here, on plugging a value of 1 to each of
a,b,c leaves us with nothing but the sum of coefficients on the RHS.
Thus, sum of coefficients = (1 + 1 +1)20
= 320
Example 4: Amit is standing at Point A and needs to reach Point B. The path exists as a 5*5 Grid and he can take only 1 step to the right, or 1 step front up. Amit's 1 step equals length/width of each grid cell. In how many ways can he reach Point B.
Amit needs to take 10 steps to reach point B. Of these he needs to make 5 right steps and 5 up steps.
If,
R -> Amit takes a Right Step
U -> Amit takes an Up step.
Then coefficient of R5U5 in (R+U)10 represents the number of ways in which he can make 5R steps and 5U Steps, thus reaching B.
Coefficient = Total ways = C(10,5) = 252
Regards & Best Wishes,
-Angadbir
