Question 1: Find the number of words formed by permuting all the letters of the following words:
i) \text{INDEPENDENCE}       ii) \text{INTERMEDIATE}       iii) \text{ARRANGE}       iv) \text{INDIA}       v) \text{PAKISTAN}       vi) \text{RUSSIA}       vii) \text{SERIES} viii) \text{EXERCISES}       ix) \text{CONSTANTINOPLE}

Answer:

i) \text{INDEPENDENCE}

There are 12 letters in the word  \text{INDEPENDENCE} out of which 2 are \text{D's} , 3 are \text{N's} and 4 are \text{E's}   and the rest are all distinct.

Therefore the total number of words

= \frac{12!}{2!3!4!}

= \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4!}{2!3!4!}

= \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 }{2 \times 3 \times 2}

= 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5

= 1663200

ii) \text{INTERMEDIATE}

There are 12 letters in the word  \text{INTERMEDIATE} out of which 2 are \text{I's} , 2 are \text{T's} and 3 are \text{E's}   and the rest are all distinct.

Therefore the total number of words

= \frac{12!}{2!2!3!}

= \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2!}{2!2!3!}

= \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 }{2 \times 3 \times 2}

= 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3

= 19958400

iii) \text{ARRANGE}  

There are 7 letters in the word  \text{ARRANGE} out of which 2 are \text{A's} , 2 are \text{R's} and the rest are all distinct.

Therefore the total number of words

= \frac{7!}{2!2!}

= \frac{7 \times 6 \times 5 \times 4 \times 3 \times 2!}{2!2!}

= \frac{7 \times 6 \times 5 \times 4 \times 3 }{2 \times 1}

= 7 \times 6 \times 5 \times 2 \times 3

= 1260    

iv) \text{INDIA}

There are 5 letters in the word  \text{INDIA} out of which 2 are \text{I's} and the rest are all distinct.

Therefore the total number of words

= \frac{5!}{2!}

= \frac{5 \times 4 \times 3 \times 2!}{2!}

=  5 \times 4 \times 3 

= 60    

v) \text{PAKISTAN}

There are 8 letters in the word  \text{PAKISTAN} out of which 2 are \text{A's} and the rest are all distinct.

Therefore the total number of words

= \frac{8!}{2!}

= \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2!}{2!}

=  8 \times 7 \times 6 \times 5 \times 2 \times 3

= 20160

vi) \text{RUSSIA}

There are 6 letters in the word  \text{RUSSIA} out of which 2 are \text{S's} and the rest are all distinct.

Therefore the total number of words

= \frac{6!}{2!}

= \frac{6 \times 5 \times 4 \times 3 \times 2!}{2!}

=  6\times 5 \times 4 \times 3 

= 360    

vii) \text{SERIES}

There are 6 letters in the word  \text{SERIES} out of which 2 are \text{S's} , 2 are \text{E's} and the rest are all distinct.

Therefore the total number of words

= \frac{6!}{2! 2!}

= \frac{6 \times 5 \times 4 \times 3 \times 2!}{2! 2!}

= \frac{6\times 5 \times 4 \times 3}{2} 

=  6\times 5 \times 2 \times 3 

= 180  

viii) \text{EXERCISES}

There are 9 letters in the word  \text{EXERCISES} out of which 3 are \text{E's} , 2 are \text{S's} and the rest are all distinct.

Therefore the total number of words

= \frac{9!}{3! 2!}

= \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2!}{3! 2!}

= \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 }{3 \times 2}

= 9 \times 8 \times 7 \times 6 \times 5 \times 2

= 30240    

ix) \text{CONSTANTINOPLE}

There are 14 letters in the word  \text{CONSTANTINOPLE} out of which 2 are \text{O's} , 3 are \text{N's} and 2 are \text{T's}   and the rest are all distinct.

Therefore the total number of words

= \frac{14!}{2!3!2!}

= \frac{14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4!}{2!3!2!}

= \frac{14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 }{3!}

= 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 1 \times 5

= 605404800

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Question 2: In how many ways can the letters of the word \text{'ALGEBRA'} be arranged without changing the relative order of the vowels and consonants?

Answer:

Number of consonants in \text{'ALGEBRA'}  = 4

Number of vowels in \text{'ALGEBRA'}  = 3 of which 2 are \text{A's}

Therefore the number of ways you can arrange the consonants = ^4P_4 = \frac{4!}{0!} = 4!

And the number of ways you can arrange the vowels = \frac{3!}{2!} = 3

Hence the total number of ways the letters of the word \text{'ALGEBRA'} be arranged without changing the relative order of the vowels and consonants = 4! \times 3 = 24 \times 2 = 72

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Question 3: How many words can be formed with the letters of the word \text{'UNIVERSITY'} , the vowels remaining together?

Answer:

There are 10 letters in the word  \text{UNIVERSITY} out of which 2 are \text{I's} and the rest are all distinct.

There are 4 vowels in the word  \text{UNIVERSITY} out of which 2 are \text{I's} and the rest are all distinct.

Considering all vowels as one, 7 letters can be arranged in = ^7P_7 = \frac{7!}{0!} = 7! ways

The vowels themselves can be arranged in \frac{4!}{2!} = 4 \times 3 = 12 ways.

Hence the number of words can be formed with the letters of the word \text{'UNIVERSITY'} , the vowels remaining together = 12 \times 7! = 60480

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Question 4: Find the total number of arrangements of the letters in the expression a^3 b^2 c^4 when written at full length.

Answer:

There are 9 letters in the in the expression a^3 b^2 c^4 out of which 3 are \text{a's} , 2 are \text{b's} , 4 are \text{c's} .

Therefore the total number of expressions

= \frac{9!}{3!2!4!}

= \frac{ 9 \times 8 \times 7 \times 6 \times 5 \times 4!}{3!2!4!}

= \frac{ 9 \times 8 \times 7 \times 6 \times 5 }{6 \times 2}

= 9 \times 4 \times 7 \times 1 \times 5 

= 1260

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Question 5: How many words can be formed with the letters of the word \text{'PARALLEL' } so that all \text{L's} do not come together?

Answer:

There are 8 letters in the word  \text{PARALLEL} out of which 2 are \text{A's} , 3 are \text{L's} and the rest are all distinct.

Therefore the total number of words possible

= \frac{8!}{2!3!}

= \frac{ 8 \times 7 \times 6 \times 5 \times 4 \times 3!}{2!3!}

= \frac{ 8 \times 7 \times 6 \times 5 \times 4 }{ 2}

= 8 \times 7 \times 6 \times 5 \times 2  

= 3360

Now let us find the number of words where \text{L's} are together.

Considering all \text{L's} as one, 6 letters, including 2 are \text{A's} can be arranged in

= \frac{6!}{2!}

= \frac{ 6 \times 5 \times 4 \times 3 \times 2!}{2!}

= 6 \times 5 \times 4 \times 3

= 360

Therefore, the number of words that can be formed with the letters of the word \text{'PARALLEL' } so that all \text{L's} do not come together = 3360-360 = 3000

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Question 6: How many words can be formed by arranging the letters of the word \text{'MUMBAI' } so that all \text{M's } come together?

Answer:

There are 6 letters in the word  \text{MUMBAI} out of which 2 are \text{M's} and the rest are all distinct.

Considering all \text{M's} as one, 5 letters can be arranged in = ^57P_5 = \frac{5!}{0!} = 5! = 120 ways

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Question 7: How many numbers can be formed with the digits 1, 2, 3, 4,3,2,1 so that the odd digits always occupy the odd places?

Answer:

Total number of digits = 7

\fbox{1} \ \fbox{2} \ \fbox{3} \ \fbox{4} \ \fbox{5} \ \fbox{6} \ \fbox{7}

Number of odd digits = 4 \text{ { 1, 3, 3, 2} }

Number of odd places = 4

The odd digits can be arranged in = \frac{4!}{2!2!} = 3 \times 2 = 6

The rest of the three even digits 2,2,4 can be arranged in = \frac{3!}{2!} = 3 

Hence the total number of arrangements = 6 \times 3 = 18

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Question 8: How many different signals can be made from 4 red, 2 white and 3 green flags by arranging all of them vertically on a flagstaff?

Answer:

Total number of flags = 9

Number of red flags = 4

Number of white flags = 2

Number of green flags = 3

Total number of signals that can be created

= \frac{9!}{4!2!3!}

= \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4!2!3!}

= \frac{9 \times 8 \times 7 \times 6 \times 5}{2 \times 6}

= 9 \times 4 \times 7 \times 1 \times 5

= 1260

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Question 9: How many number of four digits can be formed with the digits 1, 3,3,0 ?

Answer:

Total number of digits = 4

Number of 4 digit numbers that can be formed = \frac{4!}{2!} = 12

However, 0 cannot be in the first place.

Number of three digit numbers that can be formed = \frac{3!}{2!} = 3

Therefore the number of four digits can be formed with the digits 1, 3,3,0  = 12 - 3 = 9

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Question 10: In how many ways can the letters of the word \text{'ARRANGE' } be arranged so that the two \text{R's } are never together?

Answer:

There are 7 letters in the word  \text{ARRANGE} out of which 2 are \text{A's} , 2 are \text{R's} and the rest are all distinct.

Therefore the total number of words possible

= \frac{7!}{2!2!}

= 7 \times 6 \times 5 \times 1 \times 3 \times 2 \times 1

= 1260

Now let us find the number of words where \text{R's} are together.

Considering all \text{R's} as one and we have 2 are \text{A's} can be arranged in

= \frac{6!}{2!}

= 6 \times 5 \times 4 \times 3 \times 1 \times 1

= 360

Therefore, the number of words that can be formed with the letters of the word \text{'ARRANGE' } so that all \text{L's} do not come together = 1260-360 = 900

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Question 11: How many different numbers, greater than 50000 can be formed with the digits 0,1,1,5,9 .

Answer:

Total number of digits = 5

For the number to be greater then 50000 , the first digit should be either 5 or 9 .

Number arrangements with 5 in the first place = \frac{4!}{2!} = 12

Number arrangements with 9 in the first place = \frac{4!}{2!} = 12

Therefore the different numbers, greater than 50000 can be formed with the digits 0,1,1,5,9  = 12 + 12 = 24

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Question 12: How many words can be formed from the letters of the word \text{'SERIES' } which start with \text{S } and end with \text{S } ?

Answer:

\fbox{S} \ \fbox{ 2  } \ \fbox{3 } \ \fbox{4 } \ \fbox{5 } \ \fbox{S} 

We have to fill in the middle 4 boxes.

Therefore the number of words that can be formed from the letters of the word \text{'SERIES' } which start with \text{S } and end with \text{S } = \frac{4!}{2!} = 12

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Question 13: How many permutations of the letters of the word \text{'MADHUBANI' } do not begin with \text{M } but end with \text{I } ?

Answer:

Total number of letters in \text{'MADHUBANI' }  = 9

Total number of words that can be made from \text{'MADHUBANI' }   ending with \text{I } = \frac{8!}{2!} = 8 \times 7 \times 6 \times 5 \times 4 \times 3 = 20160

Number of word starting with \text{M } and ends with \text{I } = \frac{7!}{2!} = 7 \times 6 \times 5 \times 4 \times 3 = 2520

Therefore the number of words that do not begin with \text{M } but end with \text{I } = 20160-2520 = 17640

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Question 14: Find the number of numbers, greater than a million, that can be formed with the digits 2,3, 0,3, 4,2,3 .

Answer:

Total number of digits = 7

Total number of numbers that can be made from 7 digits

= \frac{7!}{2! 3!} = \frac{7 \times 6 \times 5 \times 4 \times 3!}{2! 3!} = 7 \times 6 \times 5 \times 2= 420

If 0 is at the first position. Total number of numbers that can be made from 6 digits = \frac{6!}{2! 3!} = \frac{ 6 \times 5 \times 4 \times 3!}{2! 3!} = 6 \times 5 \times 2 = 60

Hence the total number of numbers, greater than a million, that can be formed with the digits 2,3, 0,3, 4,2,3  = 420 - 60 = 360

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Question 15: There are three copies each of 4 different books. In how many ways can they be arranged in a shelf?

Answer:

There are three copies each of 4 different books

Total number of books = 12

Hence the number of possible arrangements = \frac{12!}{3!3!3!3!}

= \frac{12 \times 11 \times 10 \times 9 \times 8 \times7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 6 \times 6 \times 6 }

= 2 \times 11 \times 10 \times 1 \times 8 \times7 \times 1 \times 5 \times 1 \times 3 \times 2 \times 1

= 369600

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Question 16: How many different arrangements can be made by using all the letters in the word \text{'MATHEMATICS' } . How many of them begin with \text{C } ? How many of them begin with \text{T } ?

Answer:

There are 11 letters in the word  \text{ARRANGE} out of which 2 are \text{M's} , 2 are \text{A's} , 2 are \text{T's} and the rest are all distinct.

Therefore the number of words that can be made = \frac{11!}{2!2!2} = 4989600

If \text{C's} is the first character, then the number of words that can be made = \frac{10!}{2!2!2} = 453600

If \text{T's} is the first character, then the number of words that can be made = \frac{10!}{2!2!2} = 453600

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Question 17: A biologist studying the genetic code is interested to know the number of possible arrangements of 12 molecules in a chain. The chain contains 4 different molecules represented by the initials A (for Adenine), C (for Cytosine), G (for Guanine) and T (for Thymine) and 3 molecules of each kind. How many different such arrangements are possible?

Answer:

Total number of molecules = 12

The chain contains 4 different molecules represented by the initials A (for Adenine), C (for Cytosine), G (for Guanine) and T (for Thymine) and 3 molecules of each kind.

Therefore the total number of possible arrangements = \frac{12!}{3! 3! 3! 3!}

= \frac{12 \times 11 \times 10 \times 9 \times 8 \times7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 6 \times 6 \times 6 }

= 2 \times 11 \times 10 \times 1 \times 8 \times7 \times 1 \times 5 \times 1 \times 3 \times 2 \times 1

= 369600

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Question 18: In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row if the discs of the same color are indistinguishable?

Answer:

Total number of discs = 9

We have 4 red, 3 yellow and 2 green discs

Therefore the total number of possible arrangements = \frac{9!}{4! 3! 2!}

= \frac{9 \times 8 \times7 \times 6 \times 5 \times 4! }{4! 3! 2!}

= \frac{9 \times 8 \times7 \times 6 \times 5 }{6 \times 2}

= 9 \times 4 \times7 \times 1 \times 5

= 1260

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Question 19: How many numbers greater than 1000000 can be formed by using the digits 1, 2, 0, 2, 4, 2, 4 ?

Answer:

Total number of digits = 7

Total number of numbers that can be made from 7 digits

= \frac{7!}{2! 3!} = \frac{7 \times 6 \times 5 \times 4 \times 3!}{2! 3!} = 7 \times 6 \times 5 \times 2= 420

If 0 is at the first position.

Total number of numbers that can be made from 6 digits = \frac{6!}{2! 3!} = \frac{ 6 \times 5 \times 4 \times 3!}{2! 3!} = 6 \times 5 \times 2 = 60

Hence the total number of numbers, greater than a 1000000, that can be formed with the digits 1, 2, 0, 2, 4, 2, 4  = 420 - 60 = 360

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Question 20: In how many ways can the letters of the word \text{'ASSASSINATION '} be arranged so that all the \text{S's } are together?

Answer:

There are 13 letters in the word  \text{ASSASSINATION} out of which 3 are \text{A's} , 4 are \text{S's} , 2 are \text{I's} , 2 are \text{N's} and the rest are all distinct.

Considering all \text{S's} together, we need to arrange 10 letters.

Hence the 10 letters can be arranges in = \frac{10!}{3! 2! 2!}

= \frac{10 \times 9 \times 8 \times7 \times 6 \times 5 \times 4 \times 3! }{3! 2! 2!}

= \frac{10 \times 9 \times 8 \times7 \times 6 \times 5 \times 4 }{ 2! 2!}

= 10 \times 9 \times 8 \times7 \times 6 \times 5 \times 1

= 151200

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Question 21: Find the total number of permutations of the letters of the word \text{ 'INSTITUTE' } ?

Answer:

There are 9 letters in the word  \text{INSTITUTE} out of which 2 are \text{I's} , 3 are \text{T's}    and the rest are all distinct.

Total number of permutations of the letters of the word \text{ 'INSTITUTE' }  = \frac{9!}{2! 3!}

= \frac{ 9 \times 8 \times7 \times 6 \times 5 \times 4 \times 3! }{ 2! 3!}

= \frac{10 \times 9 \times 8 \times7 \times 6 \times 5 \times 4 }{ 2!}

= 10 \times 9 \times 8 \times7 \times 6 \times 5 \times 2

= 302400

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Question 22: The letters of the word \text{'SURITI'} are written in all possible orders and these words are written out as in a dictionary. Find the rank of the word \text{'SURITI' } .

Answer:

In a dictionary, the words are arranged in alphabetical order.

Distinct letters in \text{'SURITI' } are \text{S, U, R, I, T}

If you arrange them in alphabetical order then we have \text{I, R , S, T, U}

Number of words that start with \text{I} = 5! = 120

Number of words that start with \text{R} = \frac{5!}{2!} = 60

Now we need to find words that start with S and are before \text{'SURITI' }

These would be words starting with \text{SI, SR, ST, SUI and SURI } .

Number of words that start with \text{SI} = 4! = 24

Number of words that start with \text{SR} = \frac{4!}{2!} = 12

Number of words that start with \text{ST} = \frac{4!}{2!} = 12

Number of words that start with \text{SUI} = 3! = 6

Number of words that start with \text{SUR} = \frac{3!}{2!} = 3   ( this includes \text{'SURITI' } ) This will also contain \text{'SURTII' } which will be after \text{'SURITI' } in the dictionary. Hence you have to subtract it.

Number of words that start with \text{SURI} = 2! = 2 ( this includes \text{'SURITI' } )

Hence the rank of \text{'SURITI' }  = 120+60+24+12+12+6+(3-2) + ( 2-1) = 236

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Question 23: If the letters of the word \text{'LATE' } be permuted and the words so formed be arranged as in a dictionary, find the rank of the word \text{LATE } .

Answer:

In a dictionary, the words are arranged in alphabetical order.

Distinct letters in \text{LATE } are \text{L, A, T, E}

If you arrange them in alphabetical order then we have \text{A, E, L, T}

Number of words that start with \text{A} = 3! = 6

Number of words that start with \text{E} = 3! = 6

Number of words that start with \text{L} = 3! = 6  But one of the words itself is \text{LAET and LATE}

Hence the rank = 6 + 6 + 2= 14

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Question 24: If the letters of the word \text{'MOTHER' } are written in all possible orders and these words are written out as in a dictionary, find the rank of the word \text{'MOTHER' } .

Answer:

In a dictionary, the words are arranged in alphabetical order.

Distinct letters in \text{MOTHER } are \text{M, O, T, H, E, R}

If you arrange them in alphabetical order then we have \text{E, H, M, O, R, T}

Number of words that start with \text{E} = 5! = 120

Number of words that start with \text{H} = 5! = 120

Number of words that start with \text{ME} = 4! = 24

Number of words that start with \text{MH} = 4! = 24

Number of words that start with \text{MOE} = 3! = 6

Number of words that start with \text{MOH} = 3! = 6

Number of words that start with \text{MOR} = 3! = 6

Number of words that start with \text{MOT} = 3! = 6 However, in this combination, there will be the following words MOTHER, MOTHRE, MOTEHR, MOTERH, MOTRHE, MOTREH. Of which one is MOTHER and MOTHRE, MOTRHE, MOTREH will come after MOTHER. Hence the total number to count there is only 3 .

Hence the rank of MOTHER is = 5! + 5! + 4! + 4!  + 3! + 3! + 3! + 3 = 309

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Question 25: If the permutations oi \text{a, b, c, d, e} taken all together be written down in alphabetical order as in dictionary and numbered, find the rank of the permutation \text{debac} .

Answer:

Number of words that start with \text{a} = 4! = 24

Number of words that start with \text{b} = 4! = 24

Number of words that start with \text{c} = 4! = 24

Number of words that start with \text{da} = 3! = 6

Number of words that start with \text{db} = 3! = 6

Number of words that start with \text{dc} = 3! = 6

Number of words that start with \text{de} that we need to consider = 3 \text{ deabc, deacb, debac}

Therefore the number of words after which you will reach \text{debac} = 24 + 24 + 24 + 6 + 6 + 6 + 3 = 93

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Question 26: Find the total number of ways in which six '+' and four '-' signs can be arranged in a line such that no two '-' signs occur together.

Answer:

Number of '+' sign = 6

Number of '-' sign = 4

6 '+' signs can be arranged in \frac{6!}{6!} = 1 way.

Between the '+' signs, there are 7 places where '-' sign can go.

Therefore the 4 identical '-' signs can be arranged in \frac{^7P_4}{4!} = \frac{\frac{7!}{4!}}{4!} = 35

Hence the total number of ways to arrange the signs = 1 \times 35 = 35 

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Question 27: In how many ways can the letters of the word \text{'INTERMEDIATE' } be arranged so that:

(i) the vowels always occupy even places?
(ii) the relative order of vowels and consonants do not alter?

Answer:

There are 12 letters in the word  \text{INTERMEDIATE} out of which 2 are \text{I's} , 2 are \text{T's} , 3 are \text{E's}   and the rest are all distinct.

i)       Six vowels can be arranged in six even places  = \frac{6!}{2! 3!} = 60

The remaining six consonants can be arranged in \frac{6!}{2!} = 360

Therefore the total possible arrangements = 60 \times 360 = 21600

ii)     The relative order of vowels and consonants do not alter.

Arranging six vowels without disturbing their relative places = \frac{6!}{2! 3!} = 60

Similarly, arranging six consonants without disturbing their relative places = \frac{6!}{2!} = 360

Therefore the total possible arrangements = 60 \times 360 = 21600

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Question 28: The letters of the word \text{'ZENITH' } are written in all possible orders. How many words are possible if all these words are written out as in a dictionary ? What is the rank of the word , \text{ZENITH' } ?

Answer:

In a dictionary, the words are arranged in alphabetical order.

Distinct letters in \text{ZENITH } are \text{Z, E, N, I, T, H}

If you arrange them in alphabetical order then we have \text{E, H, I, N, T, Z}

Number of words that start with \text{E} = 5! = 120

Number of words that start with \text{H} = 5! = 120

Number of words that start with \text{I} = 5! = 120

Number of words that start with \text{N} = 5! = 120

Number of words that start with \text{T} = 5! = 120

Number of words that start with \text{ZEH} = 3! = 6

Number of words that start with \text{ZEI} = 3! = 6

Number of words that start with \text{ZENH} = 2! = 2

Number of words that start with \text{ZENI} = 2! = 2

Therefore the rank of the word , \text{ZENITH' }  = 120 + 120 + 120 + 120 + 120 + 6 + 6 + 2 + 2 = 616