Class 11: Permutations – Exercise 16.4 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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 6 ways.

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

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

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

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

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

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

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 $\text{'SUNDAY'} = 6$

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

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

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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 $\text{SUNDAY} = 6$

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

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

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

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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 $\displaystyle = ^9P_9 = \frac{9!}{0!} = 9!$

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

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

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 $\displaystyle = ^5P_4 = \frac{5!}{1!} = 5!$

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 $\displaystyle = ^6P_6 = \frac{6!}{0!} = 6! = 720$

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

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

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

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 $\displaystyle = ^3P_3 = \frac{3!}{0!} = 3!$

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 $\displaystyle = ^3P_3 = \frac{3!}{1!} = 3!$

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 game of lawn tennis mixed double be made up of 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 $\displaystyle = ^7P_2 \times ^5P_2 = \frac{7!}{5!} \times \frac{5!}{3!} = \frac{7!}{3!} = 7 \times 6 \times 5 \times 4 = 840$

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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 $\displaystyle \frac{m! (m+1)!}{(m-n+1)!}$

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 $\displaystyle = m! \times ^{m+1}P_n = \frac{m!(m+1)!}{(m+1-n)!}$

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