Fun with words (Part 1)

Probability Level 5

$\begin{aligned} \text{Number of A's} &=& 6 \\\text{Number of B's}&=& 6 \\ \text{Number of C's} &=&6 \\\text{Number of D's} &=& 5 \\\text{Number of E's} &=& 5 \\\text{Number of F's} &=& 4 \\\text{Number of G's} &=& 4 \\\text{Number of H's} &=& 3 \\ \text{Number of I's} &=& 3 \end{aligned}$

Find the number of 5 lettered words which you can make from the above given set of letters .

Details And Assumptions:

Please note that the following numbers represent the number of that particular letter available for word making . As an explicit example you have 6 A's , 6 B's etc available with you for making the word.

The answer is 58965.

2 solutions

Sankush Gupta
Dec 19, 2015

by principle of inclusion exclusion every letter of the five letter word can be chosen in 9 ways total no. of ways=9^5 but we would have to exclude the cases where there are 4 or more H or I's and 5 or more F or G's. total no. of such cases=(2x8x5!/4!)+4=84 therefore answer=9^5-84 =58965

Nice one. but will look good with latex.

the beauty of the problem is to think of inclusion exclusion as the number of letters available to us are nearby to 5 only.so unfavourable cases will be quite less . otherwise as in the other part of the problem using inclusion exclusion will create a mess.

Prakhar Bindal - 5 years, 5 months ago

true.i would advise all to try part 2 too.that is more interesting.

sankush gupta - 5 years, 5 months ago

Gopinath No
Jun 13, 2016

Using the exponential generating function:

$G(x) = \left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}\right)^3 \left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}\right)^2\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}\right)^2\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}\right)^2$

the solution is $\left[\frac{x^5}{5!}\right]G(x) = \frac{3931}{8}\times 5! = 58965$

Fun with words (Part 1)

The answer is 58965.

2 solutions

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