Permutation-Combination
Aptitude

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Q.

If the letters of the word $CHASM$ are rearranged to form $5$ letter words such that none of the word repeat and the results arranged in ascending order as in a dictionary what is the rank of the word $CHASM$?

 A.

24

 B.

31

 C.

32

 D.

30

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Solution:
Option(C) is correct

The $5$ letter word can be rearranged in $5! = 120$ Ways without any of the letters repeating.

The first $24$ of these words will start with $A$.

Then the $25^{th}$ word will start will $CA \_ \_ \_.$

The remaining $3$ letters can be rearranged in $3! = 6$ Ways. i.e. $6$ words exist that start with $CA$.

The next word starts with $CH$ and then $A$, i.e., $CHA \_ \_$.

The first of the words will be $CHAMS$. The next word will be $CHASM$.

Therefore, the rank of $CHASM$ will be,

$24 + 6 + 2$

$= 32$

Edit: For an alternative solution, check comment by Dake.


(5) Comment(s)


Stolyar
 ()

ACHMS is the first word. CHMS are arranged in 4!=24 ways, so

CAHMS is the twenty-fifth word. HMS are arranged in 3!=6 ways, so

CHAMS is the thirty-first word

CHASM is the thirty-second



Chandni
 ()

ACHMS is the first word. CHMS are arranged in 4!=24 ways, so

CAHMS is the twenty-fifth word. HMS are arranged in 3!=6 ways, so

CHAMS is the thirty-first word

CHASM is the thirty-second



Dake
 ()

In the beginnings, all the letters in this word are "remaining" (C, H, A, S, M).

C is the first letter. There is only one letter (A) among the remaining ones (H, A, S, M), which is earlier than C.

$\dfrac{1 \times 5!}{5} +$

H is the second letter. There is only one letter (A) among the remaining ones (A, S, M), which is earlier than H.

$+\dfrac{1 \times 4!}{4}+$

A is the third letter. There are no letters among the remaining ones (S, M), which are earlier than A.

$+\dfrac{0 \times 3!}{3}+$

S is the fourth letter. There is only one letter (M) among the remaining ones (M), which is earlier than S.

$+\dfrac{1 \times 2!}{2}$

This is how many permutations of CHASM are alphabetical before CHASM. So, the rank of CHASM is the sum of the previous number plus one. $\dfrac{5!}{5}+\dfrac{4!}{4}+\dfrac{2!}{2}+1$

$=\textbf{32}$.



Risham
 ()

couldn't understand



Gopal
 ()

cant understand