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Since we know, 

Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is \(\frac{n!}{p!\times q!\times r!}\).

i.e.,

The number of repeated objects of same type are in denominator multiplication with factorial.

Given, 

The word CONSTANTINOPLE. It has 14 letters and it has 3 repeated letters ‘O’, ‘N,’ and ‘T.’ The letter O is repeated twice, the letter N is repeated thrice, and letter T is repeated twice. And all other letters are distinct.

The problem can now be rephrased as to find total number of permutations of 14 objects (C, O, N, S, T, A, N, T, I, N, O, P, L, E) of which two objects are of same type (O, O), three objects are of another type (N, N, N), and the two objects are of different type (T, T).

Total number of such permutations = \(\frac{14!}{2!\times 3!\times 2!}\) 

= \(\frac{14!}{24}\) 

= 3632428750 

Hence, 

Total number of permutations of the word CONSTANTINOPLE is 3632428750.

(i) RUSSIA

Here's are 6 letters in the word ‘RUSSIA’ out of which 2 are S’s and the rest all are distinct.

Therefore by using the formula,

n!/ (p! × q! × r!)

The total number of arrangements = 6! / (2!)

= [6 × 5 × 4 × 3 × 2 × 1] / 2!

= 6 × 5 × 4 × 3

= 360

(ii) SERIES

There are 6 letters in the word ‘SERIES’ out of which 2 are S’s, 2 are E’s and the rest all are distinct.

Therefore by using the formula,

n!/ (p! × q! × r!)

The total number of arrangements = 6! / (2! 2!)

= [6 × 5 × 4 × 3 × 2 × 1] / (2! 2!)

= 6 × 5 × 3 × 2 × 1

= 180

(iii) EXERCISES

There are 9 letters in the word ‘EXERCISES’ out of which 3 are E’s, 2 are S’s and the rest all are distinct.

Therefore by using the formula,

n!/ (p! × q! × r!)

The total number of arrangements = 9! / (3! 2!)

= [9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1] / (3! 2!)

= [9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1] / (3 × 2 × 1 × 2 × 1)

= 9 × 8 × 7 × 5 × 4 × 3 × 1

= 30240

(iv) CONSTANTINOPLE

Here's 14 letters in the word ‘CONSTANTINOPLE’ out of which 2 are O’s, 3 are N’s, 2 are T’s and the rest all are distinct.

Therefore by using the formula,

n!/ (p! × q! × r!)

The total number of arrangements = 14! / (2! 3! 2!)

= 14!/ (2 × 1 × 3 × 2 × 1 × 2 × 1)

= 14! / 24

(i) INDEPENDENCE

Here's 12 letters in the word ‘INDEPENDENCE’ out of which 2 are D’s, 3 are N’s, 4 are E’s and the rest all are distinct.

Therefore by using the formula,

n!/ (p! × q! × r!)

The total number of arrangements = 12! / (2! 3! 4!)

= [12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1] / (2! 3! 4!)

= [12 × 11 × 10 × 9 × 8 × 7 × 6 × 5] / (2 × 1 × 3 × 2 × 1)

= 11 × 10 × 9 × 8 × 7 × 6 × 5

= 1663200

(ii) INTERMEDIATE

Here's 12 letters in the word ‘INTERMEDIATE’ out of which 2 are I’s, 2 are T’s, 3 are E’s and the rest all are distinct.

Therefore by using the formula,

n!/ (p! × q! × r!)

The total number of arrangements = 12! / (2! 2! 3!)

= [12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1] / (2! 2! 3!)

= [12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 3 × 2 × 1] / (3!)

= 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5

= 19958400

(iii) ARRANGE

There are 7 letters in the word ‘ARRANGE’ out of which 2 are A’s, 2 are R’s and the rest all are distinct.

Therefore by using the formula,

n!/ (p! × q! × r!)

The total number of arrangements = 7! / (2! 2!)

= [7 × 6 × 5 × 4 × 3 × 2 × 1] / (2! 2!)

= 7 × 6 × 5 × 3 × 2 × 1

= 1260

(iv) INDIA

Here's 5 letters in the word ‘INDIA’ out of which 2 are I’s and the rest all are distinct.

Therefore by using the formula,

n!/ (p! × q! × r!)

The total number of arrangements = 5! / (2!)

= [5 × 4 × 3 × 2 × 1] / 2!

= 5 × 4 × 3

= 60

(v) PAKISTAN

Here's 8 letters in the word ‘PAKISTAN’ out of which 2 are A’s and the rest all are distinct.

Therefore by using the formula,

n!/ (p! × q! × r!)

The total number of arrangements = 8! / (2!)

= [8 × 7 × 6 × 5 × 4 × 3 × 2 × 1] / 2!

= 8 × 7 × 6 × 5 × 4 × 3

= 20160

Given as

The word ‘GANESHPURI’

Here 10 letters in the word ‘GANESHPURI’. Total number of words formed is ¹⁰P₁₀ = 10!

(i) the letter G always occupies the first place?

If we fix the first position with letter G, now remaining number of letters is 9.

Number of arrangements of 9 things, taken all at a time is ⁹P₉ = 9! Ways.

Thus, a possible number of words using letters of ‘GANESHPURI’ starting with ‘G’ is 9!

(ii) the letter P and I respectively occupy the first and last place?

If we fix the first position with letter P and I in the end, now remaining number of letters is 8.

Number of arrangements of 8 things, taken all at a time is ⁸P₈ = 8! Ways.

Thus, a possible number of words using letters of ‘GANESHPURI’ starting with ‘P’ and ending with ‘I’ is 8!

(iii) Are the vowels always together?

Here 4 vowels and 6 consonants in the word ‘GANESHPURI’.

Considering 4 (A,E,I,U) vowels as one letter, now total number of letters is 7 (A,E,I,U, G, N, S, H , P, R)

Number of arrangements of 7 things, taken all at a time is ⁷P₇ = 7! Ways.

(A, E, I, U) can be put together in 4! Ways.

Thus, total number of arrangements in which vowels come together is 7! × 4!

(iv) the vowels always occupy even places?

The number of vowels in the word ‘GANESHPURI’ = 4(A, E, I, U)

The number of consonants = 6(G, N, S, H, R, I)

The even positions are 2, 4, 6, 8 or 10

Then, we have to arrange 10 letters in a row such that vowels occupy even places. Here 5 even places (2, 4, 6, 8 or 10). 4 vowels can be arranged in these 5 even places in ⁵P₄ ways.

The remaining 5 odd places (1, 3, 5, 7, 9) are to be occupied by the 6 consonants in ⁶P₅ ways.

Therefore, by using the formula,

P (n, r) = n!/(n – r)!

P (5, 4) × P (6, 5) = 5!/(5 – 4)! × 6!/(6 – 5)!

= 5! × 6!

Thus, number of arrangements therefore that the vowels occupy only even positions is 5! × 6!

Given as

The word ‘SUNDAY’

The total number of letters in the word ‘SUNDAY’ is 6.

Therefore, number of arrangements of 6 things, taken all at a time is ⁶P₆

= 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

Then, we shall find the number of words starting with N

Therefore let’s fix the first position with letter N, now remaining number of letters is 5.

Number of arrangements of 5 things, taken all at a time is ⁵P₅ = 5! = 5 × 4 × 3 × 2 × 1 = 120

Then, we need to find out a number of words starting with N and ending with Y

Therefore let’s fix the first position with letter N and Y at the end, now remaining number of letters is 4 which can be arranged in ⁴P₄ ways. = 4! = 4 × 3 × 2 × 1 = 24

Thus, the total number of words can be made by letters of the word ‘SUNDAY’ is 720.

Possible number of words using letters of ‘SUNDAY’ starting with ‘N’ is 120.

Possible number of words using letters of ‘SUNDAY’ starting with ‘N’ and ending with ‘Y’ is 24.

To find: number of words starting with C and end with Y 

There are 8 letters in word COURTESY. 

Here the position of the letters C and Y are fixed which is 1^st and 8^th .

- - - - - - 

C ? ? ? ? ? Y

Rest 6 letters are to be arranged in 6 places which can be done in P(6,6).

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)! 

Therefore, a permutation of 6 different objects in 6 places is

P(6,6) \(\frac{6!}{(6-6)!}\) = \(\frac{6!}{0!}\) = \(\frac{720}{1}\)  = 720. 

Therefore, total number of words starting with C and ending with Y is 720.

To find: 4 lettered word from letters of word NUMBERS 

There are 7 alphabets in word NUMBERS.

The word is a 4 different letter word. 

Formula

 Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)

Therefore, a permutation of 7 different objects in 4 places is 

P(7,4) = \(\frac{7!}{(7-4)!}\) = \(\frac{7!}{3!}\) = \(\frac{5040}{6}\) = 840. 

Hence, they can be arranged in 840 words.

There are 6 letters in the word SUNDAY. 

Different words formed using 6 letters of the word SUNDAY is P(6,6)

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is 

P(n,r) = n!/(n-r)! 

Therefore, a permutation of 6 different objects in 6 places is 

P(6,6) = \(\frac{6!}{(6-6)!}\) = \(\frac{6!}{0!}\) = \(\frac{720}{1}\)  = 720. 

720 words can be formed using letters of the word SUNDAY. 

When a word begins with D.

Its position is fixed, i.e. the first position.

Now rest 5 letters are to be arranged in 5 places.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, a permutation of 5 different objects in 5 places is 

P(5,5) = \(\frac{5!}{(5-5)!}\) = \(\frac{5!}{0!}\) = \(\frac{120}{1}\) = 120

Therefore, the total number of words starting with D are 120.

5 boys can be seated in 5! ways to sit in a row. Now, we can place 3 girls in 4 empty seats between them so that each girl is between 2 boys, and this can be done in ⁴P₃ = 24 ways.

As, the operations are dependent, so, the number of ways in which 5 boys and 3 girls can be seated in a row so that each girl is between 2 boys.

= 5! \(\times\)24 

= 2880.

The discussion can be shown pictorially as :

[X = 4 empty seats between 2 boys(B)]

Given: We have 5 boys and 3 girls 

To Find: Number of ways of seating so that 5 boys and 3 girls are seated in a row and each girl is between 2 boys 

The formula used: The number of permutations of n different objects taken r at a time 

(object does not repeat) is ⁿP_r =  \(\frac{n!}{(n-r)!}\)

The only arrangement possible is 

B__B__B__B__B 

Number of ways for boys =  ⁿP_r

= ⁵P₅

= \(\frac{5!}{(5-5)!}\) 

= \(\frac{5!}{0!}\)

= 120 

There are 3 girls, and they have 4 vacant positions 

Number of ways for girls = ⁴P₃ = 24 ways

= \(\frac{4!}{(4-3)!}\)  

= \(\frac{4!}{1!}\) 

= 24

Total number of ways = 24 × 120 = 2880

In 2880 ways 5 boys and 3 girls can be seated in a row so that each girl is between 2 boys.

Given: We have 3 digits, i.e. 0, 1 and 2 

To Find: Number of 5-digit numbers formed 

Let us represent the arrangement

For forming a 5-digit number, we have to fill 5 vacant spaces. 

But the first place cannot be filled with 0, hence for filling first place, we have only 1 and 2 

First place can be filled with 2 numbers, i.e. 1, 2 = 2 ways

Second place can be filled with 3 numbers = 3 ways 

Third place can be filled with 3 numbers = 3 ways 

The fourth place can be filled with 3 numbers = 3 ways 

The fifth place can be filled with 3 numbers = 3 ways 

Total number of ways = 2 × 3 × 3 × 3 × 3 = 162 

162 5-digit numbers can be formed by using the digits 0, 1 and 2.

To find: number of arrangements of 6 women drawing water from 6 wells

Here, 6 wells are needed to be used by 6 women. 

Therefore any one of the 6 women can draw water from the 1 well. 

Similarly, any 5 women can draw water from the 2nd well and so on. Lastly, there will be single women left to draw water from the 6th well.

Formula: 

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)! 

Therefore, permutation of 6 different objects in 6 places is

P(6,6) = \(\frac{6!}{(6-6)!}\)

= \(\frac{6!}{0!}\) = \(\frac{720}{1}\)

Hence, this can be done in 720 ways.

Given: We have 4 women and 4 tap

To Find: Number of ways of drawing water 

Condition: No tap remains unused 

Let us represent the arrangement

The first woman can use any of the four taps = 4 Ways 

The second woman can use the remaining three taps = 3 ways 

The third woman can use the remaining two taps = 2 ways 

The fourth woman can use the remaining one tap = 1 way 

Total number of ways = 4 × 3 × 2 × 1 = 24 

There is 24 number of ways in which 4 women can draw water from 4 taps such that no tap remains unused.

1^st woman can draw water from anyone of the 4 taps. 

Now, as, it is given, that, 

No tap should remain unused 

So, 

2^nd woman can draw water from any of the remaining 3 taps. 

Similarly, 

3^rd women can draw water from any of the remaining 2 taps. 

And, 

4^th woman have to draw water from the remaining 1 tap. 

As, 

The operations are dependent, 

So, 

Total number of ways = 4\(\times\)3\(\times\)2\(\times\)1

= 4! 

Alternative Approach : 

The number of permutations of 4 objects taken 4 at a time is = ⁴P₄ 

= 4!.

Total possible outcomes in a throw of 3 dice = 6³ = 216. 

When, 

No die shows an even number i.e. all the die shows odd number (1, 3 or 5), 

Then, 

The number of possible outcomes = 3³ = 27. 

Hence, 

The total number of possible outcomes in a throw of 3 dice in which at least one of the dice shows an even number = 216 - 27 

= 189.

To find: the value of ⁹P₀

Formula Used: 

Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by,

ⁿP_r = \(\frac{n!}{(n-r)!}\)

Therefore, 

⁹P₀ = \(\frac{6!}{(6-6)!}\)

⁹P₀ = 1 

Thus, the value of ⁹P₀ is 1.

Given, 

The molecules’ initials A, G, T, and C (All are repeated thrice). 

AAAGGGTTTCCC 

To find : Number of arrangements of these 12 molecules in such a way that all arrangements must be distinct. 

The problem can now be rephrased as to find a number of permutations of 12 objects in which 3 objects are of one type, 3 objects are of another type, 3 objects are of a third type, and remaining 3 objects are of different type. 

Since we know,

Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n! 

And, 

We also know,

Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is \(\frac{n!}{p!\times q!\times r!}\).

i.e.,

The number of repeated objects of same type are in denominator multiplication with factorial.

The number of permutation of 12 objects with repeating molecules in the factor of 3 = \(\frac{12!}{3!\times 3!\times 3!\times 3!}\) 

= 369600 

Hence,

Total number of permutation of given 12 molecules will be equals to 369600.

Given, 

The disc : Red, Yellow, and Green. 

Total of 9 discs and it has 3 repeated discs Green, Red, and Yellow. 

The disc Green is repeated twice, and disc Red is repeated 4 times, disc Yellow is repeated thrice. 

To find : Number of the arrangement of discs. 

The problem can now be rephrased as to find total number of permutations of 9 objects (R, R, R, R, Y, Y, Y, G, G) of which three objects are of same type (Y, Y, Y), and two objects are of another type (G, G), and 4 objects are of different type (R, R, R, R). 

Since we know,

Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n! 

And, 

We also know,

Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is \(\frac{n!}{p!\times q!\times r!}\).

i.e., 

The number of repeated objects of same type are in denominator multiplication with factorial.

Total number of such permutations will be \(\frac{9!}{4!\times 2!\times 3!}\).

= \(\frac{5\times 6\times 7\times 8\times 9}{(2\times 1)\times (3\times 2\times 1)}\)

= 1260 

Hence, 

Total number of permutations of the given type is 1260.

To find: no of ways in which the balls can be arranged in a row where some balls are of the same kind 

Total number of balls = 3 + 4 + 5 = 12

3 are of 1 kind, 4 are of another kind, 5 are of the third kind 

Number of ways = \(\frac{12!}{3!4!5!}\) = 27720

They can be arranged in 27720 ways

3 out of 10 members in a family know driving.

∴ On the driver’s seat a member can be place in ³P₁ ways.

Now, there are 5 seats in a car including the driver’s seat.
The remaining 4 members out of remaining 9 members of the family can be placed in ⁹P₄ ways.

∴ Total permutations of such arrangement

= ³P₁ × ⁹P₄

= 3 × (9 × 8 × 7 × 6)

= 3 × 3024

= 9072

⁹P₉ = \(\frac{9!}{(9−9)!}\)

= \(\frac{9!}{0!}\)

= 9!

= 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 362880

∴ ⁹P₉ = 362880

Using all the digits 1, 2, 3, 0, 7, 9, six digit numbers are to be formed.
∴ Excluding digit 0, one of the five digits can be placed at the first place in ⁵P₁ ways

Now, remaining 5 digits (including 0) can be arranged in remaining 5 places in ⁵P₅ ways.

∴ Total permutations for six digit numbers

= ⁵P₁ × ⁵P₅

= 5 × 5!

= 5 × 120

= 600

Hence, 600 numbers of 6 digit can be formed.

(i) 1/4! + 1/5! = x/6!

As we know that

5! = 5 × 4 × 3 × 2 × 1

6! = 6 × 5 × 4 × 3 × 2 × 1

Therefore by using these values,

1/4! + 1/5! = x/6!

1/4! + 1/(5×4!) = x/6!

(5 + 1) / (5×4!) = x/6!

6/5! = x/(6×5!)

x = (6 × 6 × 5!)/5!

= 36

Thus, the value of x is 36.

(ii) x/10! = 1/8! + 1/9!

As we know that

10! = 10 × 9!

9! = 9 × 8!

Therefore by using these values,

x/10! = 1/8! + 1/9!

x/10! = 1/8! + 1/(9 × 8!)

x/10! = (9 + 1)/(9 × 8!)

x/10! = 10/9!

x/(10 × 9!) = 10/9!

x = (10 × 10 × 9!)/9!

= 10 × 10

= 100

∴ The value of x is 100.

(iii) 1/6! + 1/7! = x/8!

As we know that

8! = 8 × 7 × 6!

7! = 7 × 6!

Therefore by using these values,

1/6! + 1/7! = x/8!

1/6! + 1/(7 × 6!) = x/8!

(1 + 7)/(7 × 6!) = x/8!

8/7! = x/8!

8/7! = x/(8 × 7!)

x = (8 × 8 × 7!)/7!

= 8 × 8

= 64

Hence, the value of x is 64.

Given as

n!(n + 2) = n! + (n + 1)!

Let us consider as RHS = n! + (n + 1)!

n! + (n + 1)! = n! + (n + 1) (n + 1 – 1)!

= n! + (n + 1)n!

= n!(1 + n + 1)

= n!(n + 2)

= L.H.S

L.H.S = R.H.S

Thus, Proved.

Given Equation : 

(n+2)! = 2550 x n!

 To Find : Value of n 

Formula : n! = n x (n-1)!

By given equation,

(n + 2)! = 2550 × n!

By using above formula we can write, 

∴ (n + 2) × (n + 1) × (n!) = 2550 × n! 

Cancelling the term (n)! from both the sides, 

∴ (n + 2) × (n + 1) = 2550 

∴ (n + 2) × (n + 1) = (51) × (50) 

Comparing both the sides, we get, 

∴n = 49 

Conclusion : Value of n is 49.

Note : Instead of taking product of two brackets in eq(1), it is easy to convert the constant term that is 2550 into product of two consecutive numbers and then by observing two sides of equation we can get value of n.

Given Equation : 

(n+3)! = 56 x (n+1)!

To Find : Value of n 

Formula : n! = n x (n-1)!

By given equation,

(n + 3)! = 56 × (n + 1)! 

By using above formula we can write, 

∴ (n + 3) × (n + 2) × (n + 1)! = 56 × (n + 1)! 

Cancelling the term (n + 1)! from both the sides, 

∴ (n + 3) × (n + 2) = 56 

∴ (n + 3) × (n + 2) = (8) × (7) 

Comparing both the sides, we get,

∴ n = 5

Conclusion : Value of n is 5.

Note : Instead of taking product of two brackets in eq(1), it is easy to convert the constant term that is 56 into product of two consecutive numbers and then by observing two sides of equation we can get value of n.

Given : 

(n + 3)! = 56 [(n + 1)!] 

To Find : n 

⇒ (n + 3) × (n + 2) × (n + 1)! = 56 [(n + 1)!] 

⇒ (n + 3) × (n + 2) = 56 

⇒ (n + 3) × (n + 2) = 8 × 7 

⇒ n + 3 = 8 

n = 5 

Hence, 

n = 5

Given as

1/9! + 1/10! + 1/11! = 122/11!

Now, let us consider LHS: 1/9! + 1/10! + 1/11!

1/9! + 1/10! + 1/11! = 1/9! + 1/(10×9!) + 1/(11×10×9!)

= (110 + 11 + 1)/(11 × 10 × 9!)

= 122/11!

= RHS

Thus proved.

(i) 30!/28!

Now, let us evaluate,

30!/28! = (30 × 29 × 28!)/28!

= 30 × 29

= 870

(ii) (11! – 10!)/9!

Now, let us evaluate,

As we know that,

11! = 11 × 10 × 9 × …. × 1

10! = 10 × 9 × 8 × … × 1

9! = 9 × 8 × 7 × … × 1

On using these values we get,

(11! – 10!)/9! = (11 × 10 × 9! – 10 × 9!)/9!

= 9!(110 – 10)/9!

= 110 – 10

= 100

(iii) L.C.M. (6!, 7!, 8!)

Now, let us find the LCM of (6!, 7!, 8!)

As we know that,

8! = 8 × 7 × 6!

7! = 7 × 6!

6! = 6!

Therefore,

L.C.M. (6!, 7!, 8!) = LCM [8 × 7 × 6!, 7 × 6!, 6!]

= 8 × 7 × 6!

= 8!

First letter can be posted in any of the 5 letter boxes i.e. in 5 ways. 

Similarly, 

Each one of the other 3 letters can be posted in any of the 5 letter boxes i.e. in 5 ways. 

As, 

The operations are dependent, 

So, 

Total number of ways = 5\(\times\)5\(\times\)5\(\times\)5

= 5⁴.

Each letter has 4 possible letter boxes option. 

So the number of ways in which 5 letters can be posted in 4 letter boxes =4 × 4 × 4 × 4 × 4 = 4⁵ (Each 4 for each letter.)

We have to find the possible number of ways in which we can post five letters among seven letter boxes when repetition of distribution of letters is allowed. 

We will use the concept of multiplication because there are five sub jobs dependent on each other and are performed one after the other. 

The thing that is distributed is considered to have choices, not the things to which we have to give them; it means that in this problem the letters have choices more precisely seven choices are there for each letter and letter boxes won’t choose any because letters have the right to choose. 

The number of ways in which we can post letters among seven letter boxes where repetition of distribution is allowed 7 × 7 × 7 × 7 × 7 

= 7⁵ 

= 16807.

According to the given question one parcel can be posted in 5 ways, 

That is in either of the one post offices so ⁵C₁. 

Similarly,

For other parcels also it can be posted in ⁵C₁ ways. 

Hence,

The number of ways the parcels be sent by registered post is ⁵C₁ × ⁵C₁ × ⁵C₁ × ⁵C₁ 

= 5 × 5 × 5 × 5 

= 625

Given, 

The word SURITI. 

Arranging the permutations of the letters of the word SURITI in a dictionary : 

To find : Rank of word SURITI in dictionary. 

First comes, words starting with letter I = 5! = 120 

Words starting from letter R = 5!/2! = 60 

Words starting from SI = 4! = 24 (4 letters no repetation) 

Words starting from SR = 4!/2! = 12 (4 letters, one repetation of I) 

Words starting from ST = 4!/2! = 12 (4 letters, one repetation of I) 

Words starting from SUI = 3! = 6 (3 letters no repetation) 

Words starting from SUR; SURIIT = 1 

SURITI = 1 

Rank of the word SURITI = 120 + 60 + 24 + 12 + 12 + 6 + 1 + 1 

= 236 

Hence,

The rank of the word SURITI in arranging the letters of SURITI in a dictionary among its permutations is 236.

Each AP consists of a unique first term and common difference. 

Number of ways to select the first term of a given set is ³C₁ = 3 

Number of ways to select a common difference of given set is ⁵C₁ = 5 

Hence,

Total number of AP’s possible are ³C₁ × ⁵C₁ 

= 3 × 5 

= 15

Given,

The total number of letters in ‘TRIANGLE’ = 8 

To find : number of words, with or without meanings using all the letters of the word like triangel or angletri etc. 

Formula used : 

Number of arrangements of n things taken all at a time = P(n, n)

P(n, r) = \(\frac{n!}{(n-r)!}\)

∴ The total number of ways in which this can be done 

= the number of arrangements of 8 things taken all at a time 

= P(8, 8)

= \(\frac{8!}{(8-8)!}\)

= \(\frac{8!}{0!}\) 

{∵ 0! = 1} 

= 8! 

Hence, 

Number of words, with or without meanings using all the letters of the word ‘TRIANGLE’ are 8!.

Option : (D)

As, It is required that, two particular persons A and B there are always two persons so, let us consider this arrangement be "AxxB" and consider it as a single object. 

So, 

We are left with, 4 persons and an object, i.e. total 5 objects. 

Now, 

This 5 objects can be arranged in 5! ways. 

Again, 

The two 'x' are to be filled with 2 persons from 6 persons, this can be done in ⁶P₂ = 30 ways. 

Two persons ‘A’ and ‘B’ can be arranged in 2! = 2 ways. 

So, 

The total number of different ways in which 8 persons can stand in a row so that between two particular persons A and B there are always two persons, is = 5! x 30 x 2

= 5! x 60.

Given : There are two works each of 3 volumes and two works each of 2 volumes.

To find : Number of ways in which these books can be arranged in a shelf provided volumes of the same work are not separated.

Let w₁, w₂, w₃, w₄, are four works 

w₁ has n₁, n₂, n₃ as volumes 

w₂ has m₁, m₂, m₃ as volumes 

w₃ has a₁, a₂ as volumes 

w₄ has b₁, b₂ as volumes 

Now,

Firstly we have to arrange these 4 works like w₂ w₃ w₁ w₄ or w₁ w₂ w₄ w₃ 

This can be done in 4! ways 

Now,

We have to separately arrange volumes of these 4 works w₁ has 3 volumes which can be arranged like n₂ n₁ n₃ or n₃ n₁ n₂ 

Volumes of w₁ can be arranged in 3! ways 

Similarly, 

Volumes of w₂ can be arranged in 3! ways Volumes of w₃ can be arranged in 2! ways 

Volumes of w₄ can be arranged in 2! Ways 

∴ Total number of ways = 4! × 3! × 3! × 2! × 2! 

= 24 × 6 × 6 × 2 × 2 

= 3456 

Hence,

The total number of ways in which these 10 books be placed on a shelf so that the volumes of the same work are not separated are 3456

To find: number of possibilities of a selection of answers. 

Each item in column A can select another item in column B. 

Therefore the question involves selecting each item from column A to each item in column B. this can be done in P(6,6) 

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is 

P(n,r) = n!/(n-r)! 

Therefore, a permutation of 6 different objects in 6 places is 

P(6,6) = \(\frac{6!}{(6-6)!}\)

= \(\frac{6!}{0!}\)  = \(\frac{720}{1}\) 

= 720. 

Therefore, the possible number of selecting an incorrect or correct answer is 720.

Given : The number of items in column A = 6 and in column B = 6 

A student is asked to match each item in column A with an item in column B 

To find : Possible number of correct or incorrect answers which he can give 

Let the items of column A are fixed i.e. they arrangement is not changing

Now, 

We just have to arrange items of column B

Formula used : 

Number of arrangements of n things taken all at a time = P(n, n)

 P(n,r) =  \(\frac{n!}{(n-r)!}\) 

∴ The total number of ways in which this can be done 

= the number of arrangements of 6 things taken all at a time 

= P(6, 6)

= \(\frac{6!}{(6-6)!}\)

= \(\frac{6!}{0!}\)

{∵ 0! = 1} 

= 6 × 5 × 4 × 3 × 2 × 1 

= 720 

Hence, 

Possible number of correct or incorrect answers which a student can give are 720.