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To find: types of February calendars that can be prepared. 

There are two factors to develop FEBRUARY metallic calendars 

(1) The day on the start of the year of which possibility = 7 

(2) Whether the year is leap year or not of which possibility is = 2 

So, number of FEBRUARY calendars possibilities to serve in future years

= 7 x 2 = 14

Given: 36 teachers are there in a school. 

To find: Number of ways in which one principal and one vice-principal can be appointed. 

There are 36 options of appointing principal and 35 option of appointing vice-principal since same teacher cannot be appointed as principal and vice-principal. 

Total number of ways = 36 x 35 = 1260

The number of arrangements of ‘n’ things taken all at a time = P (n, n)

Therefore by using the formula,

On using the formula,

P (n, r) = n!/(n – r)!

Total number of ways in which five children can stand in a queue = the number of arrangements of 5 things taken all at a time = P (5, 5)

Therefore,

P (5, 5) = 5!/(5 – 5)!

= 5!/0!

= 5! [Since, 0! = 1]

= 5 × 4 × 3 × 2 × 1

= 120

Thus, number of ways in which five children can stand in a queue are 120.

Given as

Total number of teachers in a school = 36

As we know that, number of arrangements of n things taken r at a time = P(n, r)

On using the formula,

P (n, r) = n!/(n – r)!

∴ Total number of ways in which this can be done = the number of arrangements of 36 things taken 2 at a time = P(36, 2)

P (36, 2) = 36!/(36 – 2)!

= 36!/34!

= (36 × 35 × 34!)/34!

= 36 × 35

= 1260

Thus, number of ways in which one principal and one vice-principal are to be appointed out of total 36 teachers in school are 1260.

To find: number of words 

Condition: consonants occupy odd places 

There are total of 7 letters in the word FAILURE. 

There are 3 consonants, i.e. F, L, R which are to be arranged in 4 places. 

The rest 5 letters can be arranged in 4! Ways. 

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, the total number of words are

P(4,3) x4! = \(\frac{4!}{(4-3)!}\times4!\) = \(\frac{4!}{1!}\times4!\) = \(\frac{24}{1}\times24\) = 576

Hence total number of arrangements is 576.

To find: number of words 

Condition: vowels should never occur together. 

There are 6 letters in the word GOLDEN in which there are 2 vowels. 

Total number of words in which vowels never come together = 

Total number of words – total number of words in which the vowels come together.

A total number of words is 6! = 720 words.

Consider the vowels as a group.

Hence there are 5 groups that can be arranged in P(5,5) ways, and vowels can be arranged in P(2,2,) ways. 

Formula:

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

P(n,r) = n!/(n-r)! 

Total arrangements = P(5,5) × P(2,2) = \(\frac{5!}{(5-5)!}\times\frac{2!}{(2-2)!}\)

= \(\frac{5!}{0!}\times\frac{2!}{0!}\) = 120 × 2 = 240.

Hence a total number of words having vowels together is 240. 

Therefore, the number of words in which vowels don’t come together is 720 – 240 = 480 words.

(i) There is no restriction on letters 

The word VOWELS contain 6 letters. 

The permutation of letters of the word will be 6! = 720 words. 

(ii) Each word begins with 

Here the position of letter E is fixed. 

Hence, the rest 5 letters can be arranged in 5! = 120 ways. 

(iii) Each word begins with O and ends with L 

The position of O and L are fixed.

Hence the rest 4 letters can be arranged in 4! = 24 ways.

(iv) All vowels come together 

There are 2 vowels which are O, E. 

Consider this group. 

Therefore, the permutation of 5 groups is 5! = 120 

The group of vowels can also be arranged in 2! = 2 ways.

Hence the total number of words in which vowels come together are 120 x 2 = 240 words. 

(v) All consonants come together 

There are 4 consonants V,W,L,S. consider this a group. 

Therefore, a permutation of 3 groups is 3! = 6 ways. 

The group of consonants also can be arranged in 4! = 24 ways. 

Hence, the total number of words in which consonants come together is 6 x 24 = 144 words.

To find: number of words

Condition: vowels occupy odd positions. 

There are 7 letters in the word MACHINE out of which there are 3 vowels namely A C E. 

There are 4 odd places in which 3 vowels are to be arranged which can be done P(4,3). 

The rest letters can be arranged in 4! ways 

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, the total number of words is

P(4,3)4!× = \(\frac{4!}{(4-3)!}\times4\)!

= \(\frac{4!}{1!}\times4!\) = \(\frac{24}{1}\times24\)  = 576. 

Hence the total number of word in which vowel occupy odd positions only is 576.

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

We will use the concept of multiplication because there are seven 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 four 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 four letter boxes where repetition of distribution is allowed 

4 × 4 × 4 × 4 × 4 × 4 × 4 

= 4⁷.

(i). We have to find the possible number of ways in which we can give four prizes among five students when no boy gets more than one price which means that there is no repetition. 

We will use the concept of multiplication because there are four 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 prizes have choices more precisely first prize will have five choices, second prize will have four choices and the choices will keep on decreasing by one as we go on giving prizes, and students won’t choose any because prizes will have the right to choose. 

The number of ways in which we can give four prizes among five students where repetition of distribution is not allowed 5 × 4 × 3 × 2 = 5! = 120.

(ii). We have to find the possible number of ways in which we can give four prizes among five students when any student can get any number of prices which means that there is a repetition of prizes. 

We will use the concept of multiplication because there are four 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 prizes have choices more precisely five choices are there for each prize and students won’t choose any because prizes have the right to choose. 

The number of ways in which we can give four prizes among five students where repetition of distribution is allowed 

5 × 5 × 5 × 5 

= 5⁴ 

= 625.

(iii). We have to find the possible number of ways in which we can give four prizes among five students when no student gets all the prizes. 

We will use the concept of multiplication because there are four 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 prizes have choices more precisely five choices are there for each prize and students won’t choose any because prizes have the right to choose. 

The number of ways in which we can give four prizes among five students when no student gets all the prices 

= number of ways in which any student can get any number of prices – the number of ways in which one student gets all the prizes. 

= 625 – 5 

= 620. 

There are five students, so any one student will get all the prizes hence there are five choices to give all the prizes to any one student.

Option (i) and (ii) both are false

Proofs :

For option (i),

L.H.S. = (2 + 3)! = (5!) = 120

R.H.S. = (2!) + (3!) = 2 + 6 = 8 

∴ L.H.S. ≠R.H.S. 

For option (ii), 

L.H.S. = (2 × 3)! = (6!) = 720 

R.H.S = (2!) × (3!) = 4 × 6 = 24 

∴L.H.S. ≠R.H.S. 

Important Notes : for any two whole numbers a and b,

(a+b)! ≠ (a!) + (b!)

(axb)! ≠ (a!) x (b!)

To find : Number of ways in which 4 people can be seated in a row containing 5 seats. 

The possible number of ways in which 4 people be seated in a row containing 5 seats = ⁷P₄

 (There are 5 places to be filled with 4 persons where arrangement doesn’t matter.)

\(7^P_4=\frac{7!}{(7-4)!}\)  .....\(\Big(nP_r=\frac{n!}{(n-r)!}\Big)\) 

= \(\frac{7!}{3!}\) 

=7 × 6 × 5 × 4 

= 840

Given: 10 buses running between Delhi and Agra. 

To Find: Number of ways a man can go from Delhi to Agra and return by a different bus. 

There are 10 buses running between Delhi and Agra so there are 10 different ways to go from Delhi to Agra. The man cannot return from the same bus he went so number of ways are reduced to 9. 

These second event occur in completion of first event so there are: 10 × 9=90 ways in which a man can go from Delhi to Agra and return by a different bus.

Given: 12 steamers plying between A and B.

To find: number of ways the round trip from A can be made. 

(i) The steamer which will go from A to B will be returning back, since the given condition is that same steamer should return. 

There are 12 steamers available so there are 12 different ways to make around trip between A & B if done on same steamer. 

(ii) If the return trip is done on different steamer than the once used in trip on going from A to B then the possible number of ways are:12 × 11=132. 

(11 because the once used in going from A to B cannot be used in returning hence, reduced by 1.)

Given : 

10 fountain pens, 12 ball pens, and 5 pencil 

We have to select one fountain pen from 10 fountain pens, one ball pen from 12 ball pens and one pencil from 5 pencils. 

Therefore,

The number of ways to select one fountain pen is ¹⁰C₁ and similarly the number of ways to select one ball pen is ¹²C₁ and number of ways to select one pencil from 5 pencils in ⁵C₁ 

Hence,

The number of ways to select one fountain pen, one ball pen and one pencil from a stationery shop is ¹⁰C₁ × ¹²C₁ × ⁵C₁ 

= 10 × 12 × 5 

= 600

4 couples are there. 2 persons are selected from 8 persons.

(i) Two persons selected are husband and wife:
If selected two persons are husband and wife. then they are one couple of 4 couples.
∴ Total combinations = ⁴C₁ = 4

(iii) One Is a male and the other is a female:
In 8 persons, 4 are males and 4 are females.
One male can be selected in ⁴C₁ ways andone female can be selected In ⁴C₁ ways.
∴ Total combinatíons = ⁴C₁ × ⁴C₁ = 4 × 4 = 16

(iii) One is a male and the other is a female but they are not husband and wife:
Total combinations
[Total combinations in which one is a male and one is female – Total combinations in which they are husband and wife]
= 16 – 4 = 12

To find: Number of distinct signals possible 

Total number of fags = 7 

2 are of 1 kind, 3 are of another kind, and 2 are of the 3rd kind 

⇒ Number of distinct signals = \(\frac{7!}{2!3!2!}\) = 210

Hence 210 different signals can be made

Correct option is (a) aⁿ

For the first row: 

There are 7 teachers in which the position of principal is fixed. 

Therefore, the teachers can be arranged in p(7,7) = 5040. 

For the second row: 

The tallest students are at the ends and can be arranged in 2! = 2 ways. 

Rest 18 students can be arranged in P(18,18) ways. 

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 18 different objects in 18 places is

P(18,18) = \(\frac{18!}{(18-18)!}\) 

= \(\frac{18!}{0!}=\frac{18!}{1}\) = 18!

Therefore, a total number of arrangements of the second row is 2 × 18! 

Total arrangements = 2 × 18! × 5040 = 10080 × 18! 

The total number of arrangements is 10080 × 18!

(i) two of them, Rajan and Tanvy, are always together 

Consider Rajan and Tanvy as a group which can be arranged in 2! = 2 ways. 

The 3 children with this 1 group can be arranged in 4! = 24 ways. 

The total number of possibilities in which they both come together is 2×24 = 48 ways. 

(ii) Two of them, Rajan and Tanvy, are never together 

Two of them are never together = total number of possible ways of sitting – total number of ways in which they sit together. 

A total number of possible way of arrangement of 5 students is 5! = 120 ways. 

Therefore, the total number of arrangement when they both don’t sit together is = 120 – 48 = 72.

Candidates in mathematics are not sitting together = total ways – the 

Students are appearing for mathematic sit together. 

The total number of arrangements of 8 students is 8! = 40320 

When students giving mathematics exam sit together, then consider 

Them as a group. 

Therefore, 6 groups can be arranged in P(6,6) ways. 

The group of 3 can also be arranged in 3! Ways. 

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, total arrangement are 

P(6,6) × 3! = \(\frac{6!}{(6-6)!}\times3!\) 

= \(\frac{6!}{0!}\times3!\) = \(\frac{720}{1}\times6\) = 4320

The total number of possibilities when all the students giving 

Mathematics exam sits together is 4320 ways. 

Therefore, number of ways in which candidates appearing 

Mathematics exam is 40320 – 4320 = 36000.

Option : (C)

In the room there are 12 bulbs of same wattage, each having separate switch. 

We have 2 options for each bulb i.e. either switching it ON or to switching it OFF. 

The 1^st bulb can be switched ON or OFF i.e. 2 choices. 

The 2^nd bulb can be switched ON or OFF i.e. 2 choices, and so on. 

Now, 

For 12 bulbs we have total 2¹² choices. 

But, as we have to illuminate the room, so the single choice in which all the 12 bulbs are OFF should be omitted. 

So, we have total (2¹² – 1) ways to light the room with different amounts of illumination.

To Prove : 

LCM {6!, 7!, 8!} = 8! 

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

LCM is the smallest possible number that is a multiple of two or more numbers. 

Here, we observe that (8!) is the first number which is a multiple of all three given numbers i.e. 6!, 7! and 8!.

1 x (8!) = 8!

8 x (7!) = 8!

8 x 7 x (6!) = 8!

Therefore, 8! is the LCM of {6!, 7!, 8!}

Conclusion : Hence proved

To find: number of 6 digit 

0 cannot be in the first place because that would make a 5 - digit number 

Total number of 6 - digit numbers = Total number of number possible - Number of numbers with 0 at the first place 

Total number of numbers possible = \(\frac{6!}{2!2!}\) = 180

Number of numbers with 0 at first place = \(\frac{5!}{2!2!}\) = 30

⇒ Number of 6 - digit numbers = 180 - 30 = 150 

150 six - digit numbers are possible

Given : 

(n + 1)! = 90 [(n – 1)!] 

To Find : n ? 

⇒ (n + 1) × (n) × (n-1)! 

= 60 [(n – 1)!] 

⇒ (n + 1) × (n) = 90 

⇒ (n + 1) × (n) = 10 × 9 

n = 9 

Hence, 

n = 9

Alphabetical arrangement of letters: A,D,I,N 

⇒ 1^st word: ADIIN 

To find other words: 

Case 1: words starting with A 

Number of words = \(\frac{4!}{2!}\) = 12

⇒ 13^th word starts with D and is DAIIN

Case 2: words starting with D

Number of words = \(\frac{4!}{2!}\) = 12

Case 3: Words starting with I

Number of words = 4! = 24

⇒ (12+12+24+1)^th = 49^th word starts with N and is NAIID

Case 4: Words starting with N

Number of words = \(\frac{4!}{2!}\) = 12

⇒ (48+12)^th word is the last word which starts with N 

⇒ 60^th word = NDIIA 

1^st word: ADIIN 

13^th word: DAIIN 

9^th word: NAIID

60^th word: NDIIA

Given,

(n + 2)! = 60 [(n – 1)!] 

To Find : n 

⇒ (n + 2) × (n + 1) × (n) × (n-1)! = 60 [(n – 1)!] 

⇒ (n + 2) × (n + 1) × (n) = 60 

⇒ (n + 2) × (n + 1) × (n) 

= 5 × 4 × 3 

n = 3 

Hence, n = 3

(i) (2 + 3)! = 2! + 3!

Now, let us consider LHS: (2 + 3)!

(2 + 3)! = 5!

Then RHS,

2! + 3! = (2 × 1) + (3 × 2 × 1)

= 2 + 6

= 8

LHS ≠ RHS

Thus, the given expression is false.

(ii) (2 × 3)! = 2! × 3!

Let us consider as LHS: (2 × 3)!

(2 × 3)! = 6!

= 6 × 5 × 4 × 3 × 2 × 1

= 720

Then RHS,

2! × 3! = (2 × 1) × (3 × 2 × 1)

= 12

LHS ≠ RHS

Hence, the given expression is false.

Given : 

n! (n + 2) = n! + (n + 1)! 

R.H.S. = 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 

Hence, Proved.

Given,

3 . 6 . 9 . 12 . 15 . 18 

Find : Convert into Factorial 

= (3 × 1) × (3 × 2) × (3 × 3) × (3 × 4) × (3 × 5) × (3 × 6) 

= 3⁶ (1 × 2 × 3 × 4 × 5 × 6) 

= 3⁶ (6!)

(i) 5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 10

Now, let us evaluate

Then, we can write it as:

5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 10 = (1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10)/(1 × 2 × 3 × 4)

= 10!/4!

(ii) 3 ⋅ 6 ⋅ 9 ⋅ 12 ⋅ 15 ⋅ 18

Let us evaluate

3 ⋅ 6 ⋅ 9 ⋅ 12 ⋅ 15 ⋅ 18 = (3 × 1) × (3 × 2) × (3 × 3) × (3 × 4) × (3 × 5) × (3 × 6)

= 3⁶ (1 × 2 × 3 × 4 × 5 × 6)

= 3⁶ (6!)

(iii) (n + 1) (n + 2) (n + 3) … (2n)

Let us evaluate

(n + 1) (n + 2) (n + 3) … (2n) = [(1) (2) (3) .. (n) … (n + 1) (n + 2) (n + 3) … (2n)]/(1) (2) (3) .. (n)

= (2n)!/n!

(iv) 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 … (2n – 1)

Let us evaluate

1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 … (2n – 1) = [(1) (3) (5) … (2n - 1)] [(2) (4) (6) … (2n)] / [(2) (4) (6) … (2n)]

= [(1) (2) (3) (4) … (2n-1) (2n)] / 2ⁿ [(1) (2) (3) … (n)]

= (2n)! / 2ⁿ n!

Given : 

(2 × 3)! = 2! × 3! 

L.H.S = (2 × 3)!

= 6! 

= 6 × 5 × 4 × 3 × 2 × 1 

= 720 

R.H.S = 2! × 3! 

= 2 × 1 × 3 × 2 × 1 

= 12 

Hence, 

L.H.S ≠ R.H.S

Given : 

(2 + 3)! = 2! + 3! 

L.H.S = (2 + 3)! 

= 5! 

R.H.S = 2! + 3! 

= (2 × 1) + (3 × 2 × 1) 

= 2 + 6 

= 8 

Hence, 

L.H.S ≠ R.H.S

2 screws are defective In a box of 6 screws.

∴ Non-defective screws In a box = 6 – 2 = 4

2 non-defective screws out of 4 non-defective screws can be selected

in ⁴C₂ = \(\frac{4×3}{2×1}\) = 6 ways.

Correct option is (d) m + n

Correct option is (a) mn

3 dolls, 4 kitchen sets and 3 cars are displayed in a toy shop. 3 toys are to be selected.

(i) All are dolls : There are 3 dolls.
∴ Total combinations = ³C₃ = 1

(ii) All are different toys: One doll out of 3 dolls can be selected in ³C₁ ways and one kitchen set out of 4 sets can be selected in 4CX ways and one car out of 3 cars can be selected in ³C₁ ways.
∴ Total combinations = ³C₁ × ⁴C₁ × ³C₁ =3x4x3 = 36

(iii) Two are dolls and one is a kitchen
set: Two dolls out of 3 dolls can be selected in ³C₂ ways and one kitchen set out of 4 sets can be selected in ⁴C₁ ways.
∴ Total combinations = ³C₂ × ⁴C₁
= \(\frac{3×2}{2×1} × 4\)
= 3 × 4
= 12

In a pack of 52 cards, there are 4 cards of king and 4 cards of queen.

From well shuffled pack of 52 cards, 2 cards of queen

OR

2 cards of king can be selected in ⁴C₂ +⁴C₂ ways.

∴ Total combinations = ⁴C₂ + ⁴C₂

\(\frac{4×3}{2×1} + \frac{4×3}{2×1}\)

= 6 + 6 = 12