Fun with words (Part 2)

Probability Level 5

Number of A’s = 5 Number of B’s = 5 Number of C’s = 4 Number of D’s = 4 Number of E’s = 3 Number of F’s = 1 Number of G’s = 1 Number of H’s = 1 \begin{aligned} \text{Number of A's} &=& 5 \\\text{Number of B's}&=& 5 \\ \text{Number of C's} &=&4 \\\text{Number of D's} &=&4 \\\text{Number of E's} &=& 3 \\\text{Number of F's} &=& 1 \\\text{Number of G's} &=& 1 \\\text{Number of H's} &=& 1 \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 5 A's , 5 B's etc available with you for making the word.


The answer is 21462.

Prakhar Bindal by Prakhar Bindal , 22, India

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2 solutions

Sankush Gupta
Dec 19, 2015

This one is interesting :P If you have not done part 1 ,I would recommend you to do that first then you would understand this better. when you see the 1's in F,G and H ; the first thing that should come to your mind is making cases. so case 1:when none of F,G and H is taken total no. of ways=5^5-(4x5+3)=3102[PIE]

case 2:when one of F,G and H are taken total no. of ways=3x5x(5^4-1)=9360

case 3: when two of F,G and H are taken total no. of ways=3x5p2x5^3=7500

case 4:when all three of F,G and H are taken total no. of ways=5c2x3!x5^2=1500

therefore,answer=sum of individual answers=21462

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You should try my problem "fun with numbers" i am sure u will like it

Prakhar Bindal - 5 years, 5 months ago

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I just tried all three of them.Could do two correctly.part 2 was tricky :) nice questions

sankush gupta - 5 years, 5 months ago

Nice seems you are quite versed with PIE as i hesitate in using it ! :)

Prakhar Bindal - 5 years, 5 months ago
Gopinath No
Jun 13, 2016

Using the exponential generating function:

G ( x ) = ( 1 + x + x 2 2 ! + x 3 3 ! + x 4 4 ! + x 5 5 ! ) 2 ( 1 + x + x 2 2 ! + x 3 3 ! + x 4 4 ! ) 2 ( 1 + x + x 2 2 ! + x 3 3 ! ) ( 1 + x ) 3 G(x) = \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)\left(1+x\right)^3

the solution is [ x 5 5 ! ] G ( x ) = 3577 20 × 5 ! = 21462 \left[\frac{x^5}{5!}\right]G(x) = \frac{3577}{20}\times 5! = 21462

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