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In the question,

We have to find the possible number of 3 digit numbers formed by the numbers 0 to 9 when repetition of digits is not allowed. 

We will use the concept of multiplication because there are three sub jobs dependent on each other because a number appearing on any one place will not appear in any other place. 

In the first position from left we will have ten choices, in the second position we will have nine choices, and in the third position, we will have eight choices because repetition is not allowed and one digit is occupied in each position. 

The number of ways in which we can form three - digit numbers with the help of given data is 10 × 9 × 8 

= 720 

There will be only one correct combination out of these so the incorrect combinations will be 720 - 1 

= 719

To find: number of words such that C and T are never together 

Number of words where C and T are never the together = Total numbers of words - Number of words where C and T are together 

Total number of words = \(\frac{7!}{2!}\) = 2520

Let C and T be denoted by a single letter Z 

New word is APAINZ 

This can be permuted in \(\frac{6!}{2!}\) = 360 ways 

Z can be permuted among itself in 2 ways 

⇒ Number of words where C and T are together = 360 × 2 = 720 

⇒ Number of words where C and T are never together = 2520 - 720 = 1800 

There are 1800 words where C and T are never together

Here's 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

Now, let us consider all R’s together as one letter, there are 6 letters remaining. Out of which 2 times A repeats and others are distinct.

Therefore these 6 letters can be arranged in n!/ (p! × q! × r!) = 6!/2! Ways.

Number of words in which all R’s come together = 6! / 2!

= [6 × 5 × 4 × 3 × 2!] / 2!

= 6 × 5 × 4 × 3

= 360

Therefore, now the number of words in which all L’s do not come together = total number of arrangements – The number of words in which all L’s come together

= 1260 – 360

= 900

Thus, the total number of arrangements of word ARRANGE in such a way that not all R’s come together is 900.

Given,

The word ARRANGE. 

It has 7 letters of which 2 letters (A, R) are repeating. The letter A is repeated twice, and the letter R is also repeated twice in the given word. 

To find : Number of ways the letters of word ARRANGE be arranged in such a way that not all R’s do come together.

First, 

We find all arrangements of word ARRANGE, and then we minus all those arrangements of word ARRANGE in such a way that all R’s do come together, from it. 

This will exactly be the same as- all number of arrangements such that not all R’s do come together.

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.

A total number of arrangements of word ARRANGE : 

Total letters 7. 

Repeating letters A and R, 

The letter A repeating twice and letter R repeating twice. 

The total number of arrangements.

= \(\frac{7!}{2!\times 2!}\)

Now,

We find a total number of arrangements such that all R’s do come together.

A specific method is usually used for solving such type of problems. 

According to that, 

We assume the group of letters that remain together (here R, R) is assumed to be a single letter and all other letters are as usual counted as a single letter. 

Now,

Find a number of ways as usual; 

The number of ways of arranging r letters from a group of n letters is equals to ⁿP_r. 

And the final answer is then multiplied by a number of ways we can arrange the letters in that group which has to be stuck together in it (Here R, R).

Letters in word ARRANGE : 7 letters 

Letters in a new word : 

A, RR, A, N, G, E : 6 letters (Letter A repeated twice).

Total number of word arranging all the letters \(\frac{6!}{2!}\times\)\(\frac{2!}{2!}\)

Where second fraction 2! divided by 2! comes from arranging letters inside the group RR : 

Arrangements of two letters where all the two letters are same = 2!/2! = 1 (Obviously! You can even think of it).

Now, 

A Total number of arrangements where not all R’s do come together is equals to total arrangements of word ARRANGE minus the total number of arrangements in such a way that all R’s do come together.

= \(\frac{7!}{2!\times 2!}\) - (\(\frac{6!}{2!}\times\)\(\frac{2!}{2!}\)) 

= 900 

Hence, 

The total number of arrangements of word ARRANGE in such a way that not all R’s come together is equals to 900.

Option : (B)

We have, 

6 objects {B}, {H}, {A}, {R}, {A}, {T} and there are 2 A's. 

So, the number of words that can be formed out of the letters of the word ‘BHARAT’ is

= \(\frac{6!}{2!}\)

= 360

The number of words that can be formed out of the letters of the word ‘BHARAT’ is in which {B} and {H} come together is = 5! 

= 120. 

[considering {BH} as a single object] 

Hence, 

The number of words from the letters of the word ‘BHARAT’ in which B and H will never come together, is = (360 – 120) 

= 240.

Given, 

Expression a³b²c⁴ 

i.e. in expansion aaabbcccc. 

To find : Number of expressions that can be generated by permuting the letters of given expression aaabbcccc. 

Given expression has three repeating characters a, b, and c. 

The letter a is repeated 3 times, the letter b is repeated 2 times, and the letter c is repeated 4 times. 

So, the given problem can now be rephrased as to find a total number of arrangements of 9 objects (3+2+4) of which 3 objects are of the same type, 2 objects are of another type, and 4 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 ways of arranging 9 objects of which 3, 2, and 4 objects are of different types is equaled to 

= \(\frac{9!}{3!\times 2!\times 4!}\) 

= 124

Hence,

Number of ways of arranging the letters of word/expression aaabbcccc is equals to 124.

Option : (A)

We have, 

6 objects {C}, {H}, {E}, {E}, {S}, {E} and there are 3 E's. 

So, 

The number of words that can be formed out of the letters of the word ‘CHEESE’ is

= \(\frac{6!}{3!}\) 

= 120.

Given: We have 6 letters 

To Find: Number of ways to arrange letters P,E,N,C,I,L 

Condition: N is always next to E 

Here we need EN together in all arrangements. 

So, we will consider EN as a single letter. 

Now, we have 5 letters, i.e. P,C,I,L and ‘EN’. 

5 letters can be arranged in ⁵P₅ ways

⇒ ⁵P₅

⇒ \(\frac{5!}{(5-5)!}\) 

⇒ \(\frac{5!}{0!}\)

⇒ 120

In 120 ways we can arrange the letters of the word ‘PENCIL’ so that N is always next to E.

Given: We have 6 letters 

To Find: Number of words formed with Letter of the word ‘CHEESE.’ 

The formula used: The number of permutations of n objects, where p₁ objects are of one kind, p₂ are of the second kind, ..., p_k is of a k^th kind and the rest if any, are of a different kind is = \(\frac{n!}{p_1!P_2! ......P_k!}\)

Suppose we have these words – C,H,E₁,E₂,S,E₃ 

Now if someone makes two words as CHE₁E₃SE₂ and CHE₂E₃SE₁

These two words are different because E₁, E₂ and E₃ are different but we have three similar E’s hence, in our case these arrangements will be a repetition of same words. 

In the word CHEESE, 3 E’s are similar 

∴ n = 6, p₁ = 3

\(\frac{6!}{3!}\) = \(\frac{720}{6}\) = 120

In 120 ways the letters of the word ‘CHEESE’ can be arranged.

Given: We have 9 letters 

To Find: Number of words formed with Letter of the word ‘ALLAHABAD.’ 

The formula used: The number of permutations of n objects, where p₁ objects are of one kind, p₂ are of the second kind, ..., p_k is of a kth kind and the rest if any, are of a different kind is = \(\frac{n!}{p_1!p_2! .....p_k!}\)

‘ALLAHABAD’ consist of 9 letters out of which we have 4 A’s and 2 L’s. 

Using the above formula 

Where, 

n = 9 

p₁ = 4 

p₂ = 2

\(\frac{9!}{4!2!}\)  = 7560

7560 different words can be formed by using all the letters of the word ‘ALLAHABAD.’

Option : (C)

In a group of n distinct objects, 

The number of arrangements of 4 objects is = ⁿP₄ and the number of arrangements of 2 objects is = ⁿP₂. 

According to the given problem,

\(P^n_4\) = 12 x \(P^n_2\) 

⇒ \(\frac{n!}{(n-4)!}\) = 12 x \(\frac{n!}{(n-2)!}\)

⇒ n×(n-1)×(n-2)×(n-3) = 12×n×(n-1) 

⇒ (n-2)×(n-3) =12 = (6-2)(6-3) 

⇒ n = 6 

Hence, 

The number of objects = 6.

[Note : \(P^n_4\) = ⁿP₄ and \(P^n_2\) = ⁿP₂]

Given,

The letters a, b, c, d, and e. 

Arranging the permutations of the letters a, b, c, d, and e in a dictionary : 

To find : Rank of word debac in dictionary. 

First comes, 

Words starting from letter a = 4! = 24 

Words starting from letter b = 4! = 24 

Words starting from letter c = 4! = 24 

Words starting from letter d : 

Words starting from da = 3! = 6 

Words starting from db = 3! = 6 

Words starting from dc = 3! = 6 

Words starting from dea = 2! = 4

Words starting from deb : 

debac = 1 

Rank of the word debac = 24 + 24 + 24 + 6 + 6 + 6 + 4 + 1 

= 95 

Hence, 

The rank of the word debc in arranging the letters a, b, c, d, and e in a dictionary among its permutations is 95.

Option : (D)

Primarily, 

Excluding the 3 things which are to be included, we have to select (r - 3) things from (n - 3), this can be done in ^{n – 3}C_{r – 3} ways.

Now, 

The 3 particular things can be selected from 3 remaining things only in 1 way. 

And the selected r things can be arranged in r! ways. 

So,

The number of permutations of n different things taking r at a time when 3 particular things are to be included is = r! ^{n – 3}C_{r – 3}.

To find: 

the number of permutations of 10 objects, taken 4 at a time. 

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 = \(\frac{n!}{(n-r)!}\)

¹⁰P₄  = \(\frac{10!}{6!}\)

¹⁰P₄ = 10^{x 9x8x7}

¹⁰P₄ = 5040

Hence, the number of permutations of 10 objects, taken 4 at a time is 5040.

Given : Number of boys = 6 and number of girls = 5 

To find : Possible number of arrangements in a group photograph 

Let boys be b₁, b₂, b₃, b₄, b₅, b₆ and girls be g₁, g₂, g₃, g₄, g₅ 

Possible arrangements are :

b₁ b₂ b₃ b₅ b₆ b₄ 

g₂ g₄ g₁ g₅ g₃ 

b₂ b₁ b₅ b₃ b₄ b₆ 

g₂ g₄ g₅ g₁ g₃ 

In this arrangement, 

We are arranging boys and girls separately. 

Formula used : 

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

P(n, r) = \(\frac{n!}{(n-r)!}\)

∴ Number of ways to arrange boys,

= 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! 

= 6 × 5 × 4 × 3 × 2 × 1 

= 720

Formula used : 

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

P(n, r) = \(\frac{n!}{(n-r)!}\)

∴ Number of ways to arrange girls,

= the number of arrangements of 5 things taken all at a time.

= P(5, 5)

= \(\frac{5!}{(5-5)!}\)

= \(\frac{5!}{0!}\)

{∵ 0! = 1} 

= 5! 

= 5 × 4 × 3 × 2 × 1 

= 120 

Now, 

We will get total number of ways by multiplying their separate arrangements 

∴ Total number of ways = 720 × 120 

= 86400 

Hence,

Possible number of arrangements in which 6 boys and 5 girls can be arranged for a group photograph with provided conditions are 86400.

ⁿP₃ = 990
∴ n(n – 1)(n – 2)
∴ n(n – 1)(n – 2)= 11 × 10 × 9
∴ n (n – 1) (n – 2) = 11(11 – 1) (11 – 2)
∴ n = 11

To find: the value of ⁶²P₃

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, 

⁶²P₃ = \(\frac{62!}{(62-3)!}\)

⁶²P₃ = 62 × 61 × 60 × 59 = 226920 

Thus, the value of ⁶²P₃ is 226920.

Given,

The word ASSASSINATION. 

It has 13 letters of which repeated letters are A, S, I, and N repeating thrice, 4 times, twice and twice respectively. 

We have to find number of words that can formed in such a way that all S’s must come together. 

For example - 

ASSSSANNIITO, ININAASSSSATO are few words among them. 

Notice that all S letters comes in bunch (In bold letters). 

A specific method is usually used for solving such type of problems. 

According to that we assume the group of letters that are remains together (here S, S, S, S) are assumed to be a single letter and all other letters are as usual counted as single letter. 

Now find number of ways as usual; 

Number of ways of arranging r letters from a group of n letters is equals to ⁿP_r .

And the final answer is then multiplied by number of ways we can arrange the letters in that group which has to be sticked together in it (Here S, S, S, S). 

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 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.

Now, 

Letters of word ASSASSINATION : 13 

Now from our method,

Letters are SSSS, A, A, A, I, I, N, N, T, O. (10 letters) 

Total number of arrangements of 10 letters = \(\frac{10!}{3!\times 2!\times 2!}\)

And total number of arrangements of grouped letters (Here S, S, S, S) is \(\frac{4!}{4!}\),equals 1.

So, our final answer for arranging the letters such that all vowels sticks together equals multiplication of \(\frac{10!}{3!\times 2!\times 2!}\) and \(\frac{4!}{4!}\).

Total number of arrangements = \(\frac{10!}{3!\times 2!\times 2!}\)\(\times\)\(\frac{4!}{4!}\)

= 151200. 

Hence,

Total number of words formed during arrangement of letters of word ASSASSINATION such that all S’s remains together is equals to 151200.

(i) Given : Five-digit number is required in which the first digit cannot be zero, and the repetition of digits is not allowed. 

Assume five boxes, now the first box can be filled with one of the nine available digits, so the possibility is ⁹C₁ 

Similarly,

The second box can be filled with one of the nine available digits, so the possibility is ⁹C₁ 

The third box can be filled with one of the eight available digits, so the possibility is ⁸C₁ 

The fourth box can be filled with one of the seven available digits, so the possibility is ⁷C₁ 

The fifth box can be filled with one of the six available digits, so the possibility is 6C1 

Hence,

The number of total possible outcomes is ⁹C₁ × ₉C¹ × ⁸C₁ × ⁷C₁ × ⁶C₁ 

= 9 × 9 × 8 × 7 × 6 

= 27216 

(ii) Given : Five-digit number is required in which the first digit cannot be zero, and the repetition of digits is allowed 

Assume five boxes, now the first box can be filled with one of the nine available digits, so the possibility is ⁹C₁ 

Similarly,

The second box can be filled with one of the ten available digits, so the possibility is ¹⁰C₁ 

The third box can be filled with one of the ten available digits, so the possibility is ¹⁰C₁ 

The fourth box can be filled with one of the ten available digits, so the possibility is ¹⁰C₁ 

The fifth box can be filled with one of the ten available digits, so the possibility is ¹⁰C₁ 

Hence,

The number of total possible outcomes is ⁹C₁ × ¹⁰C₁ × ¹⁰C₁ × ¹⁰C₁ × ¹⁰C₁ 

= 9 × 10 × 10 × 10 × 10 

= 90000

We have to find the possible number of three-digit numbers that are formed with the numbers 0, 1, 3, 5, 7 when repetition of digits is allowed.

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

There are four choices for hundred’s position because there is zero also which cannot be used in hundred’s place because then our number will become a two digit number instead of a three digit number, there are five choices for the ten’s place because there are a total of five numbers, hundred's placed in which zero was not included but in ten’s place zero is included and in one’s place there are also five choices because repetition is allowed. 

The number of ways in which we can form three digit numbers when repetition of digits is allowed along with given numbers 4 × 5 × 5 = 100

Given : 

Four-digit number is required which is greater than 8000 

Assume four boxes, 

Now,

In the first box can either be one of the two numbers 8 or 9, so there are two possibilities which is ²C₁ 

In the second box,

The numbers can be any of the four digits left, so the possibility is ⁴C₁ 

Similarly, 

For the third box, the numbers can be any of the three digits left, so the possibility is ³C₁ 

For the fourth box, the numbers can be any of the two digits left, so the possibility is ²C₁ 

Hence,

Total number of possible outcomes is ²C₁ × ⁴C₁ × ³C₁ × ²C₁ 

= 2 × 4 × 3 × 2 

= 48

We have to find the possible number of numbers that are formed with the numbers 0, 1, 2, 3, 4, 5 which are less than 1000 when repetition of digits is allowed.

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

First we will make three-digit numbers, 

There are five choices for hundred’s position because there is zero also which cannot be used in hundred’s place because then our number will become a two digit number instead of a three digit number, there are six choices for the ten’s place because there are a total of six numbers, hundred's placed in which zero was not included but in ten’s place zero is included and in one’s place there are also six choices because repetition is allowed. 

The number of ways in which we can form three digit numbers when repetition of digits is allowed along with given numbers 5 × 6 × 6 = 180 

Secondly we will make two-digit numbers, 

There are five choices for ten’s position because there is zero also which cannot be used in ten’s place because then our number will become a one digit number instead of a two digit number, there are six choices for the one’s place because there are a total of six numbers, ten place in which zero was not included but in one’s place zero is included. 

The number of ways in which we can form two digit numbers when repetition of digits is allowed along with given numbers 5 × 6 = 30 

Thirdly we will form single digit natural numbers, 

Which are given in the question they are five in number, 

5 because zero cannot be included because we need natural numbers. 

Hence,

The total number of numbers formed which are less than 1000 and are formed by given numbers are 180+30+5 = 215.

Given :

Six persons are to be arranged in a row 

Assume six seats,

Now in the first seat, 

Any one of six members can be seated, so the total number of possibilities is ⁶C₁ 

Similarly, 

In the second seat, 

Any one of five members can be seated, so the total number of possibilities is ⁵C₁ 

In the third seat,

Any one of four members can be seated, so the total number of possibilities is ⁴C₁ 

In the fourth seat, 

Any one of three members can be seated, so the total number of possibilities is ³C₁ 

In the fifth seat,

Any one of two members can be seated, so the total number of possibilities is ²C₁ 

In the sixth seat,

Only one remaining person can be seated, so the total number of possibilities is ¹C₁ 

Hence,

The total number of possible outcomes = ⁶C₁ × ⁵C₁ × ⁴C₁ × ³C₁ × ²C₁ × ¹C₁ 

= 6 × 5 × 4 × 3 × 2 × 1 

= 720

We have to find the possible number of four digit numbers which are less than 4321 and are formed with the numbers 1, 2, 3, 4 when repetition of digits is allowed.

We will use the concept of multiplication because there are four sub jobs dependent on each other because the number 4321 cannot be exceeded and we have to look at repetitions carefully. 

First, let us consider all the cases in which we do not assign 4 on the thousand’s place which means we will only assign 1, 2, 3 only and repetition in all other places. 

The number of ways in which we can assign values to four digit number, but the thousand’s place will not be assigned 4 when repetition is allowed with the help of given data is 3 × 4 × 4 × 4 = 192. 

Now,

We will consider the case when at thousand’s place there is only 4 and nothing else. 

So there will be only one choice on thousand’s place that is 4,there will be two choices on hundred’s place because there will not be 3 and 4 because then our number will become greater than 4321 and 3 will be dealt later by fixing it in hundred’s place like 4 is fixed in this case. The rest places have all the choices. 

The number of ways in which we can assign values to the four-digit number and the thousand’s place is only assigned 4 but hundred’s place is not assigned 3 when repetition is allowed with the help of given data is 1 × 2 × 4 × 4 = 32. 

Now we will consider the case when thousand’s place is fixed by 4 and hundred’s place is fixed by 3, ten’s place is fixed by 1. 

There is one choice each on thousand’s and hundred’s place, but on ten’s place also there is one choice which is 1,so a number of choices on one's place will be four. 

The number of ways in which we can assign values to four-digit number and the thousand’s place is only assigned 4, hundred’s place is only assigned 3, ten’s place is only assigned 1 when repetition is allowed with the help of given data is 1 × 1 × 1 × 4 = 4. 

Now we will consider the case when thousand’s place is fixed by 4 and hundred’s place is fixed by 3, ten’s place is fixed by 2. 

There is one choice each on thousand’s and hundred’s place, but on ten’s place also there is one choice which is 2,so a number of choices on one's place will be one. 

So the number coming to my mind will be 4321 , which is one number. 

Hence,

Total numbers occurring are 192+32+4+1 = 229.

Given : 

Nine-digit number is required in which the first digit cannot be zero and the repetition of digits is not allowed.

Assume nine boxes, 

Now, 

The first box can be filled with one of the nine available digits, so the possibility is ⁹C₁ 

Similarly,

The second box can be filled with one of the nine available digits, so the possibility is ⁹C₁ 

The third box can be filled with one of the eight available digits, so the possibility is ⁸C₁ 

The fourth box can be filled with one of the seven available digits, so the possibility is ⁷C₁ 

The fifth box can be filled with one of the six available digits, so the possibility is ⁶C₁ 

The sixth box can be filled with one of the six available digits, so the possibility is ⁵C₁ 

The seventh box can be filled with one of the six available digits, so the possibility is ⁴C₁ 

The eighth box can be filled with one of the six available digits, so the possibility is ³C₁ 

The ninth box can be filled with one of the six available digits, so the possibility is ²C₁ 

Hence,

The number of total possible outcomes is ⁹C₁ × ⁹C₁ × ⁸C₁ × ⁷C₁ × ⁶C₁ 

= 9 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 

= 9 (9!)

In the question,

It is given that we have to find three - digit numbers with distinct digits which means the digits should be nonrepeating and all the digits should be odd means no even digit. 

Numbers by which we form the three digit numbers are 1, 3, 5, 7, 9 only the odd ones. 

We will use the concept of multiplication because we have three sub jobs and each job is dependent on the other because a number selected on hundred’s place will not appear in ones and tens place. 

The number of ways in which we can form three - digit numbers with odd digits is, 5 × 4 × 3 = 60

We can also do it by 3 × 4 × 5 = 60, there are total of 5 choices in the first digit on any place then it becomes 4 in the second digit place because some number has been placed in the first digit out of 5 , so out of rest 4 numbers one number is again consumed at second place so at the end 3 numbers are left to fit the last and third place of our 3 digit number.

Given : 

Odd number less than 1000 is required. 

In order to make the number odd, the last digit has to either of (3, 5, 7)

Assume three boxes,

In the first either of the three digits (3,5,7) can be placed, so the possibility is ³C₁ 

Case 1: Middle digit is zero 

If the middle digit is zero, number of ways of placing odd numbers on the second box = 2 

Hence, 

The total number of ways = 3 × 2 = 6 ways 

Case 2 : Middle digit is an odd number Number of ways of filling middle box = 2 

Number of ways of filling third box = 1 

Hence, 

The total number of ways = 3 × 3 

= 9 ways 

Hence,

Total number of outcomes possible = 6 + 9 

= 15 ways

To find: number of ways of winning the first three prizes. 

The first price can go to any of the 10 students. 

The second price can go to any of the remaining 9 students. 

The third price can go to any of the remaining 8 students. 

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 10 different objects in 3 places is 

P(10,3) = \(\frac{10!}{(10-3)!}\) 

= \(\frac{10!}{7!}\) = \(\frac{3628800}{5040}\) = 720. 

Therefore, there are 10 × 9 × 8 = 720 ways to win first three prizes.

Given as

An examinee can answer a question either true or false, therefore there are two possibilities.

Number of ways for an examinee to answer a set of ten true/false type questions are: ²C₁ × ²C₁ × ²C₁ × ²C₁ × ²C₁ × ²C₁ × ²C₁ × ²C₁ × ²C₁ × ²C₁ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024 ways

Total number of ways to make an attempt to open the lock is = 10 × 10 × 10 = 1000

Number of successful attempts to open the lock = 1

Number of unsuccessful attempts to open the lock = 1000 – 1 = 999

Thus, required number of possible ways to make an unsuccessful attempt to open the lock is 999.