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Choose 4 letters from the word ' ARSUN SURESH '  ?

My Approach

A--1
R--2
S--3
U--2
N--1
E--1
H--1       11 letter word .

case 1 : All 4 letters are different

case2 : 2 letters same 2 letters different

case 3 : 2 letters same , 2 letters same

case 4 : 3 letters same , 1 letter different

case 5 : all 4 letters are same

CASE 1
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There are 7 distinct letters and we have to choose 4 from them C(7,4)

CASE 2
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2R's and any 2 distinct OR    C(2,2)* C(7,2) +

2S's and any 2 distinct OR     C(3,2)* C(7,2) +   // Choose two S from three

2U's and any 2 distinct         C(2,2)* C(7,2)

CASE 3
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2R's and 2'S      C(2,2)* C(3,2) +

2S's and 2U's    C(2,2)* C(3,2) +

2U's and 2 R's   C(2,2)* C(3,2)

CASE 4
----------
There are no enough letters .  Therefore req no of ways for case4 is 0

retagged | 182 views

+1 vote
This problem can also be solved using generating functions as:

We are interested in 4 word letters, so if we represent all possibilities for occurrence of alphabets of given string using exponential generating function as:

$A = \frac{z^{0}}{0!} + \frac{z^{1}}{1!}$

$R = \frac{z^{0}}{0!} + \frac{z^{1}}{1!} + \frac{z^{2}}{2!}$

$S = \frac{z^{0}}{0!} + \frac{z^{1}}{1!} + \frac{z^{2}}{2!} + \frac{z^{3}}{3!}$

$U = \frac{z^{0}}{0!} + \frac{z^{1}}{1!} + \frac{z^{2}}{2!}$

$N = \frac{z^{0}}{0!} + \frac{z^{1}}{1!}$

$E = \frac{z^{0}}{0!} + \frac{z^{1}}{1!}$

$H = \frac{z^{0}}{0!} + \frac{z^{1}}{1!}$

Considering the product of these generating functions, $h(z)$:
$h(z) = A*R*S*U*N*E*H$

We need to extract coefficient of $z^{4}$ to get the solution. However, this way it becomes quite time consuming and error-prone to solve.
by Active (1.8k points)
0
I dont know anything about generating functions . I feel it is the most difficult part to understand . Your answer seems to be not useful for me . Could you please help me with my approach and guide me what is wrong in that ?  Thanks. :)

I apologize for interpreting the question incorrectly. Your question asks for number of 4 alphabet selections from given set of 2 words and not the number of words of length 4. So here is my take,

Case 0: 0 repetition

$\binom{7}{4}$

Case 1: 1 repetition of length 2:

$\binom{3}{1}×\binom{4}{2}$

Case 2: 2 repetitions of length 2:

$\binom{3}{2}$

Case 3: 1 repetition of length 3:

$\binom{6}{1}$

In case 2... Once say 2 R's are selected.... You have 6 distinct alphabets to select from...  so IMO $\binom{7}{2}$ is incorrect.

In case 3:....as far as I understand your reasoning.... $\binom{3}{2}$ is incorrect since  S's are indistinguishable from each other so there is only one way to select 1/2/3 S's from a set of repetitions of S.

by Active (1.8k points)
edited

+1 vote