1How many arrangements can be made of the letters of the word “ASSASSINATION”? In how many of them are the vowels always together? A$\dfrac{{13!}}{{{{\left( {4!} \right)}^2}}},\;\dfrac{{8! \times 6!}}{{{{\left( {4!} \right)}^2}}}$ B$\dfrac{{13!}}{{6! \times 7!}},\;\dfrac{{8! \times 6!}}{{{{\left( {4!} \right)}^2}}}$ C$\dfrac{{13!}}{{6! \times 7!}},\;\dfrac{{8! \times 6!}}{{6! \times 7!}}$ D$\dfrac{{13!}}{{{{\left( {4!} \right)}^2}}},\;\dfrac{{8! \times 6!}}{{6! \times 7!}}$

Answer: A

Explanation:
Total letters in the word $ASSASSINATION$ $=13$
$(SSSS), (AAA), (II), (NN), T, O$ Formula:Number of ways of arranging $n$ items of which $p$ are alike, $q$ alike and so on $ = \dfrac{{n!}}{{p! \times q! \times ...}}$
Total number of arrangements $ = \dfrac{{13!}}{{4! \times 3! \times 2! \times 2!}}$ $ = \dfrac{{13!}}{{4! \times 4 \times 3!}}$ $ = \dfrac{{13!}}{{{{\left( {4!} \right)}^2}}} \qquad (1)$
Vowels in the given word $=AAAIIO$
Let us put them in a box and name it a letter $X$.
$\boxed{AAAIIO}_{\rightarrow{X}}, S, S, S, S, N, N, T $
Now total letters $= 8$
Number of ways of arranging $8$ letters in $8$ places $ = \dfrac{{8!}}{{4! \times 2!}}$
Numebr of ways the vowels in the box arrange themselves $ = \dfrac{{6!}}{{3! \times 2!}}$
Total ways in which all vowels together $ = \dfrac{{8!}}{{4! \times 2!}} \times \dfrac{{6!}}{{3! \times 2!}}$ $ = \dfrac{{8! \times 6!}}{{{{\left( {4!} \right)}^2}}} \qquad (2)$

2In how many ways can the letters of the word ARRANGE be arranged so that two R’s are never together A900 B360 C120 D1260

Answer: A

Explanation:

$ARRANGE$ $=(AA), (RR), N, G, E$
Two R’s are never together $=$ (Total possible arrangements) $-$ (Two R’s are always together)
Total number of arrangements = $\dfrac{{7!}}{{2! \times 2!}}$ = $\dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 2}}$ $= 1260$

Arrangements with both R’s together:
Let us put both $R$'s are in a box and name it letter $X$.
$\boxed{RR}_{\rightarrow{X}}, A, A, N, G, E $
Number of ways of arranging above $6$ letters = $\dfrac{{{\text{6}}!}}{{{\text{2}}!}}$ $= 360$
(Note: Two $R$'s in the box arrange only one way)

R’s never together $= 1260 - 360 = 900$

3In how many ways can the letters of the word ARRANGE be arranged so that The Two A’s are together but not two R’s A900 B360 C240 D1260

Answer: C

Explanation:

The Two A's are together but not two R's ＝ Two A's are always together $-$ (Two R's and two A's are always together)
Arrangements with both A's together:
Let us put both $A$'s are in a box and name it letter $X$.
$\boxed{AA}_{\rightarrow{X}}, R, R, N, G, E $
Number of ways of arranging above $6$ letters ＝ $\dfrac{{{\text{6}}!}}{{{\text{2}}!}}$ ＝ $360$
(Note: Two $A$'s in the box arrange only one way)

Arrangements with both A's and R's together:
Let us put both $R$'s are in a box and name it letter $X$ and both $A$'s are in a box and name it letter $Y$
$\boxed{RR}_{\rightarrow{X}},\boxed{AA}_{\rightarrow{Y}}, N, G, E$
Number of ways of arranging above $5$ letters $=5!$ $= 120$
(Note: Two $A$'s and two $R$'s in the boxes arrange themselves only one way)

The Two A's are together but not two R's ＝ $360 - 120 = 240$

4In how many ways can the letters of the word ARRANGE be arranged so that neither Two A’s nor two R's are together A900 B360 C120 D660

Answer: D

Explanation:

Neither 2 A’s or 2R’s are together ＝ Total possible arrangements $-$ (two A’s are always together) $-$ (Two R's are always together) $+$ (Both Two A's and Two R's are always together)
Total number of arrangements ＝ $\dfrac{{7!}}{{2! \times 2!}}$ ＝ $\dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 2}}$ ＝ $1260$
Number of arrangements of A's are together ＝ $360$ ($\because $ refer to the above two questions)
Number of arrangements of R's are together ＝ $360$ ($\because $ refer to the above two questions)
Number of arrangements of both A's are together and both R's are together ＝ $120$ ($\because $ refer to the above two questions)
Neither 2 A’s or 2R’s are together ＝ $1260 - 360 - 360 + 120 = 660$

5Ten different alphabets are given. Words containing five alphabets are to be formed from them. Find the number of words which have exactly one alphabet repeats. A${}^{10}{P_5}$ B${\rm{10}}^{\rm{5}} $ C${10^5}{ - ^{10}}{P_5}$ D$58060$

Answer: D

Explanation:
We have 10 alphabets, we are to form five alphabet word, and given that repetition of exactly one alphabet is allowed. We can choose that alphabet in 10 ways. Now, following cases are possible: 1. It repeats exactly two times: $AABCD$ =
Number of ways of selecting that repeating alphabet $ = {}^{10}{C_1}$
Number of ways of selecting remaining three alphabets $ = {}^9{C_3}$
Number of ways of arranging $AABCD$ letters $ = \dfrac{{5!}}{{2!}}$
Total numbers of ways $ = {}^{10}{C_1} \times {}^9{C_3} \times \dfrac{{5!}}{{2!}}$ $= 50400$

2. It repeats exactly three times:$AAABC$
Number of ways of selecting that repeating alphabet $ = {}^{10}{C_1}$
Number of ways of selecting remaining three alphabets $ = {}^9{C_2}$
Number of ways of arranging $AAABC$ letters $ = \dfrac{{5!}}{{3!}}$
Total numbers of ways $ = {}^{10}{C_1} \times {}^9{C_2} \times \dfrac{{5!}}{{3!}}$ $= 7200$

3. It repeats exactly four times:$AAAAB$
Number of ways of selecting that repeating alphabet $ = {}^{10}{C_1}$
Number of ways of selecting remaining three alphabets $ = {}^9{C_1}$
Number of ways of arranging $AAAAB$ letters $ = \dfrac{{5!}}{{4!}}$
Total numbers of ways $ = {}^{10}{C_1} \times {}^9{C_1} \times \dfrac{{5!}}{{4!}}$ $= 450$

4. It repeats exactly five times:$AAAAA$
Number of ways of selecting that repeating alphabet $ = {}^{10}{C_1}$
Number of ways of arranging $AAAAA$ letters $ = 1$
Total numbers of ways $ = {}^{10}{C_1} \times 1$ $= 10$

Total number of words with only one alphabet repeating $= 58060$

Note:
Total number of words that can be formed if repetition is not allowed ＝ ${}^{10}{P_5}$
If repetition is allowed total number of words that can be formed will be ＝ ${\rm{10}}^{\rm{5}} $
The number of arrangements having atleast one alphabet repeating ＝ ${10^5}{ - ^{10}}{P_5}$

6How many 4 letter words may be formed by using the letters of the word 'ASSASSINATION' A916 B917 C360 D480

Answer: B

Explanation:
$(SSSS), (AAA), (II), (NN), T, O$
We have above 6 different letters.
We can choose 4 letter words in the following ways. a. All the 4 letters are different - $ABCD$
Number of ways of choosing 4 letters from 6 different letters ＝ ${}^6{C_4}$
Number of ways of arranging $ABCD$ letters ＝ $4!$
Total ways ＝ ${}^6{C_4} \times 4!$

b. 2 letters same and 2 are different - $AABC$
Number of ways of choosing the repeated letter = ${}^4{C_1}$
Number of ways of choosing remaining two letters from 5 different letters ＝ ${}^5{C_2}$
Number of ways of arranging $ABCD$ letters ＝ $ = \dfrac{{4!}}{{2!}}$
Total ways ＝ ${}^4{C_1} \times {}^5{C_2} \times \dfrac{{4!}}{{2!}}$

c. 2 letters repeat twice - $AABB$
Number of ways of choosing the two repeated letters = ${}^4{C_2}$
Number of ways of arranging $AABB$ letters ＝ $ = \dfrac{{4!}}{{2! \times 2!}}$
Total ways ＝ ${}^4{C_2} \times \dfrac{{4!}}{{2! \times 2!}}$

d. 1 letter repeat thrice - $AAAB$
Number of ways of choosing the repeated letter = ${}^2{C_1}$
Number of ways of choosing remaining letter from 5 different letters ＝ ${}^5{C_1}$
Number of ways of arranging $ABCD$ letters ＝ $ = \dfrac{{4!}}{{3!}}$
Total ways ＝ ${}^2{C_1} \times {}^5{C_1} \times \dfrac{{4!}}{{3!}}$

e. 1 letter repeat 4 times - $AAAA$
We choose the letter in only one way (only S we can choose) and we arrange them in 1 way.