Question

In how many ways can the letter of word ASSASSINATION be arranged such that
A.All the four S came together
B.All the Vowels occur together
C.All the A do not occur together

Answer 1

Count the letters which came together and then consider it as one. Then use the formula = $\dfrac{{n!}}{{{P_1}!{P_2}!{P_3}!...}}$ where P1= of which is alike, P2= of which is alike, P3= of which is alike and so on.

Complete step-by-step answer:
Since we need to assigned 4S together,
We consider 4 S as one block
So our letter become we arrange them Now
since letters are repeating
Hence we use this formula = $\dfrac{{n!}}{{{P_1}!{P_2}!{P_3}!}}$
Here,
n = letters to be arranged = 9 + 1 = 10
Since , 3A, 2I, 2N
P1 = 3 , P2 = 2, P3 = 2
So, Number of arrangement where the 3S are together $= \dfrac{{10!}}{{3!2!2!}}$
$= \dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{3!2!2!}}$
= 151200
(ii) Since we need to assigned all vowels together
We considered all vowels as one block
We use formula = $\dfrac{{n!}}{{{P_1}!{P_2}!{P_3}!}}$ $= \dfrac{{10!}}{{4! \times 2! \times 2!}}$
n = letters to be arranged = 7 + 1 = 8

Since, 4S, 2N, 1T
P1 = 4, P2 = 2, P1 = 1
So, number of arrangement where the vowels are together = $\dfrac{{8!}}{{4! \times 2! \times 1!}}$
$= \dfrac{{8 \times 7 \times 6 \times 5 \times {{4!}}}}{{{{4!}} \times 2 \times 1 \times 1}}$
= 840
(iii) Total number of permutation of all the I not coming together
= Total Permutation – Total permutation of all A coming together
Total permutation
In ASSASSINATION
there are 4S, 3A, 2I, 2N, 1O, 1T
since letter are repeating
We will use the formula = $\dfrac{{n!}}{{{P_1}!{P_2}!{P_3}!{P_4}!}}$
Total number of alphabet = 13
Hence, n = 13
Also, there are 4S, 3A, 2I, 2N
P1 = 4, P2 = 3, P3 = 2, P4 = 2
Hence,
Total number of Permutation = $\dfrac{{n!}}{{{P_1}!{P_2}!{P_3}!{P_4}!}}$
$= \dfrac{{13!}}{{4!3!2!2!}}$
$= \dfrac{{13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times {{4!}}}}{{{{4!}} \times 3 \times 2 \times 2 \times 2 \times 1}}$
= 10,810800
Total permutation of all A coming together
Now taking 3As as one,
Here, there are repeating letters
So, we use the formula, Number of Permutation = $\dfrac{{n!}}{{{P_1}!{P_2}!{P_3}!}}$
Number of letters = 10 (4S, 2I, 2N, 1O, 1T, 3A)
Since there are 4S, 2I, 2N
P1 = 4, P2 = 2, P3 = 2
Number of Permutation of all A together = $= \dfrac{{10!}}{{4! \times 2! \times 2!}}$
$= \dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times {{4!}}}}{{{{4!}} \times 2 \times 1 \times 2 \times 1}}$
= 37800
Now,
Total number of permutation of all A not coming together
= 10810800 – 37800
= 10773000

Note: In this type of Question, first assigned all the repeating letter in one block and then as per question proceed for further and use formula = $\dfrac{{n!}}{{{P_1}!{P_2}!{P_3}!}}$