Question 1: In how many ways can the letter so the word, \text{FAILURE} , be arranged so that the consonants may occupy only odd positions?

Answer:

Number of letters in \text{FAILURE}  = 7

There are 4 vowels and 3 consonants in \text{FAILURE} . In all we have 7 places to be filled.

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

Consonants can go to boxes 1, 3, 5, 7 in ^4P_3 = 4! ways

The 4 vowels can be arranged in the remaining places in ^4P_4 = 4!  ways.

Hence the total number of arrangements = 4! \times 4! = 24 \times 24 = 576

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Question 2: In how many ways can the letters of the word \text{'STRANGE} be arranged so that:

i) the vowels come together ?     ii) the vowels never come together? and     iii) the vowels occupy only the odd places?

Answer:

Number of letters in \text{'STRANGE} = 7

Number of vowels \text{( A, E)}  = 2

Number of consonants \text{( S, T, R, N, G)} = 5

i)      Considering two vowels as one, we can arrange the 6 in ^6P_6 = 6! = 720 ways.

However, the vowels themselves can be arranges in 2! ways.

Hence the total number of ways the letters can be arranges = 720 \times 2 = 1440 ways.

ii)     Total number of ways that all the 7 letter can be arranged = ^7P_7 = 7! = 5040 ways.

Therefore the total number of words where the vowels never come together = 5040  - 1440 = 3600

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

There are 4 odd places. The two vowels can be arranged in ^4P_2 = \frac{4!}{2!} = 12 ways.

The rest of the 5 consonants can be arranged in ^5P_5 = \frac{5!}{0!} = 120 ways.

Hence the total number of arrangements = 12 \times 120 = 1440  

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Question 3: How many words can be formed from the letters of the word \text{'SUNDAY'} ? How many of these begin with \text{D} ?

Answer:

Number of letters in \text{'SUNDAY'}  = 6

Therefore the total number of words that can be formed = ^6P_6 = \frac{6!}{0!} = 720

The total number of words that can be formed with D fixed at the first position = ^5P_5 = \frac{5!}{0!} = 120

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Question 4: How many words can be formed out of the letters of the word, \text{'ORIENTAL'} , so that the vowels , always occupy the odd places ?

Answer:

Number of letters in \text{'ORIENTAL'}  = 8

There are 4 vowels  in \text{'ORIENTAL'}

\fbox{1} \ \fbox{2} \ \fbox{3} \ \fbox{4} \ \fbox{5} \ \fbox{6} \ \fbox{7} \ \fbox{8}

4 vowels can go to boxes 1, 3, 5, 7 in ^4P_4 = 4! ways

The 4 consonants can be arranged in the remaining places in ^4P_4 = 4!  ways.

Hence the total number of arrangements = 4! \times 4! = 24 \times 24 = 576

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Question 5: How many words can be formed out of the letters  of the word \text{SUNDAY} ? How many of these words begin with \text{N} ? How many begin with \text{N} and end with \text{Y} .

Answer:

Number of letters in \text{SUNDAY}  = 6

Total number of words that can be formed = ^6P_6 = \frac{6!}{0!} = 720

Total number of words that can be formed with \text{N} in the first position = ^5P_5 = \frac{5!}{0!} = 120

Total number of words that can be formed with \text{N} in the first position and \text{Y} in the last position = ^4P_4 = \frac{4!}{0!} = 24

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Question 6: How many different words can be formed from the letters of the word \text{GANESHPURI} ? In how many of these words:

(i) the letter \text{G} always occupies the first place?

(ii) the letters \text{P} and \text{I} respectively occupy first and last place?

iii) the vowels are always together

(iv) the vowels always occupy even places?

Answer:

Number of letters in \text{GANESHPURI}  = 10

i)      Total number of words that can be formed with \text{G} in the first position = ^9P_9 = \frac{9!}{0!} = 9!

ii)     Total number of words that can be formed with \text{P} in the first position and \text{I} in the last position = ^8P_8 = \frac{8!}{0!} = 8!

iii)    There are 4 vowels and 6 consonants. Considering all vowels as one, we can arrange the 7 in = ^7P_7 = \frac{7!}{0!} = 7!

The 4 vowels can themselves be arranged in 4! ways.

Hence the total possible arrangements = 7! \times 4!

iv)    \fbox{1} \ \fbox{2} \ \fbox{3} \ \fbox{4} \ \fbox{5} \ \fbox{6} \ \fbox{7} \ \fbox{8} \ \fbox{9} \ \fbox{10}

There are 5 even places ( 2, 4, 6, 8, 10) . Therefore 4 vowels can be arranged in = ^5P_4 = \frac{5!}{1!} = 5!

Remaining 6 places can be occupies by the 6 consonants in 6! ways.

Hence the total number of arrangements = 5! \times 6!

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Question 7:  How many permutations can be formed by the letters of the word \text{VOWELS} when

i) there is no restriction on letters     ii) each word begins with letter \text{E}

iii) each word begins with \text{O}   and ends with \text{L}      iv) all vowels come together

v) all consonants come together

Answer:

Number of letters in \text{VOWELS}  = 6

i) Total number of words that can be formed  = ^6P_6 = \frac{6!}{0!} = 6! = 720

ii) Total number of words that can be formed with \text{E} in the first position = ^5P_5 = \frac{5!}{0!} = 5! = 120

iii) Total number of words that can be formed with \text{O} in the first position and \text{L} in the last position = ^4P_4 = \frac{4!}{0!} = 4! = 24

iv) There are 2 vowels and 4 consonants. Considering all vowels as one, we can arrange the 5 in = ^5P_5 = \frac{5!}{0!} = 5!

The 2 vowels can themselves be arranged in 2! ways.

Hence the total possible arrangements = 5! \times 2!  = 120 \times 2 = 240

v)  There are 2 vowels and 4 consonants. Considering all consonants as one, we can arrange the 3 in = ^3P_3 = \frac{3!}{0!} = 3!

The 4 consonants can themselves be arranged in 4! ways.

Hence the total possible arrangements = 3! \times 4!  = 6 \times 24 = 144

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Question 8: How many words can be formed out of letters of the word \text{'ARTICLE'} , so that vowels occupy even places.

Answer:

Number of letters in \text{'ARTICLE'}  = 7

Number of vowels = 3 \text { (A, I E) }

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

There are 3 even places ( 2, 4, 6 ) . Therefore 3 vowels can be arranged in = ^3P_3 = \frac{3!}{1!} = 3!

Remaining 4 places can be occupies by the 4 consonants in 4! ways.

Hence the total number of arrangements = 3! \times 4! = 6 \times 24 = 144

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Question 9: In how many ways can a lawn tennis mixed double be made up from seven married couples if no husband and wife play in the same set?

Answer:

We can pick 2 husbands in ^7P_2 ways. 

Now we cannot pick the two wives for the mixed doubled. We have to choose 2 wives from the 5 left in ^5P_2 ways.

Hence the total number of ways can a lawn tennis mixed double be made up from seven married couples if no husband and wife play in the same set = ^7P_2 \times ^5P_2 = \frac{7!}{5!} \times \frac{5!}{3!} = \frac{7!}{3!} = 7 \times 6 \times 5 \times 4 = 840

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Question 10: m men and n women are to be seated in a row so that no two women sit together. If m > n then show that the number of ways in which they can be seated as \frac{m! (m+1)!}{(m-n+1)!}

Answer:

m men can be seated in m! ways.

Once the men are seated, there would be (m+1) spots for the n women to be seated in. Therefore the number of ways n women can be seated in ^{m+1}P_n ways.

Hence the total number of ways m men and n women are to be seated in a row so that no two women sit together = m! \times ^{m+1}P_n = \frac{m!(m+1)!}{(m+1-n)!}

Hence proved.

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Question 11: How ‘many words (with or without dictionary meaning) can be made from the letters in the  word \text{MONDAY} , assuming that no letter is repeated, if

(i) 4 letters are used at a time?

(ii) all letters are used at a time?

(iii) all letters are used but first is vowel?

Answer:

i) Number of 4 letters words that can be made = ^6P_4 = \frac{6!}{2!} = 6 \times 5 \times 4 \times 3 = 360

ii) Number of 6 letters words that can be made = ^6P_6 = \frac{6!}{0!} = 6! = 720

iii) First letter is a vowel. There are two vowels.

Hence the number of words that can be made = 2! \times ^5P_5 = 2 \times 5! = 240

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Question 12: How many three letter words can be made using the letters of the word \text{ORIENTAL} ?

Answer:

Number of letters in \text{ORIENTAL}  = 8

Number of 3 letters words that can be made = ^8P_3 = \frac{8!}{5!} = 8 \times 7 \times 6 = 336