\displaystyle \textbf{Question 1: } \text{In how many ways can the letters of the word }\text{`FAILURE'}\text{ be}
\displaystyle \text{arranged so that the consonants occupy only odd positions?}
\displaystyle \text{Answer:}
\displaystyle \text{The word }\text{FAILURE}\text{ has }7\text{ distinct letters.}
\displaystyle \text{There are }4\text{ vowels }(A,I,U,E)\text{ and }3\text{ consonants }(F,L,R).
\displaystyle \text{The four odd positions are }1,3,5\text{ and }7.
\displaystyle \fbox{1}\quad\fbox{2}\quad\fbox{3}\quad\fbox{4}\quad\fbox{5}\quad\fbox{6}\quad\fbox{7}
\displaystyle \text{The }3\text{ consonants can be arranged in }3\text{ of the }4\text{ odd positions in }{}^4P_3\text{ ways.}
\displaystyle {}^4P_3=\frac{4!}{(4-3)!}=\frac{4!}{1!}=24.
\displaystyle \text{The }4\text{ vowels can be arranged in the remaining }4\text{ positions in }4!=24\text{ ways.}
\displaystyle \therefore \text{Required number of arrangements}=24\times24=576.
\displaystyle \\

\displaystyle \textbf{Question 2: } \text{In how many ways can the letters of the word }\text{`STRANGE'}\text{ be}
\displaystyle \text{arranged so that:}
\displaystyle \text{i) the vowels come together?}\qquad \text{ii) the vowels never come together?}
\displaystyle \text{iii) the vowels occupy only the odd positions?}
\displaystyle \text{Answer:}
\displaystyle \text{The word }\text{STRANGE}\text{ has }7\text{ distinct letters.}
\displaystyle \text{The vowels are }A,E\text{ and the consonants are }S,T,R,N,G.
\displaystyle \text{i) Consider the two vowels as one block.}
\displaystyle \text{The vowel block and the }5\text{ consonants form }6\text{ objects.}
\displaystyle \text{These }6\text{ objects can be arranged in }6!=720\text{ ways.}
\displaystyle \text{The two vowels within the block can be arranged in }2!=2\text{ ways.}
\displaystyle \therefore \text{Required number of arrangements}=6!\times2!=720\times2=1440.
\displaystyle \text{ii) Total number of arrangements of the }7\text{ letters}=7!=5040.
\displaystyle \text{Number of arrangements in which the vowels are together}=1440.
\displaystyle \therefore \text{Number of arrangements in which the vowels are not together}
\displaystyle =5040-1440=3600.
\displaystyle \text{iii) The four odd positions are }1,3,5\text{ and }7.
\displaystyle \fbox{1}\quad\fbox{2}\quad\fbox{3}\quad\fbox{4}\quad\fbox{5}\quad\fbox{6}\quad\fbox{7}
\displaystyle \text{The two vowels can be arranged in two of the four odd positions in }{}^4P_2\text{ ways.}
\displaystyle {}^4P_2=\frac{4!}{(4-2)!}=\frac{4!}{2!}=12.
\displaystyle \text{The five consonants can be arranged in the remaining five positions in }5!=120\text{ ways.}
\displaystyle \therefore \text{Required number of arrangements}=12\times120=1440.
\displaystyle \\

\displaystyle \textbf{Question 3: } \text{How many words can be formed from the letters of the word}
\displaystyle \text{`SUNDAY'? How many of these begin with }D\text{?}
\displaystyle \text{Answer:}
\displaystyle \text{The word }\text{SUNDAY}\text{ has }6\text{ distinct letters.}
\displaystyle \therefore \text{Total number of words that can be formed}={}^6P_6
\displaystyle =\frac{6!}{(6-6)!}=\frac{6!}{0!}=6!=720.
\displaystyle \text{When }D\text{ is fixed in the first position, the remaining }5\text{ letters can be arranged}
\displaystyle \text{in }{}^5P_5=\frac{5!}{(5-5)!}=\frac{5!}{0!}=5!=120\text{ ways.}
\displaystyle \therefore \text{The number of words beginning with }D\text{ is }120.
\displaystyle \\

\displaystyle \textbf{Question 4: } \text{How many words can be formed from the letters of the word}
\displaystyle \text{`ORIENTAL' so that the vowels always occupy the odd positions?}
\displaystyle \text{Answer:}
\displaystyle \text{The word }\text{ORIENTAL}\text{ has }8\text{ distinct letters.}
\displaystyle \text{The vowels are }O,I,E,A\text{ and the consonants are }R,N,T,L.
\displaystyle \text{The four odd positions are }1,3,5\text{ and }7.
\displaystyle \fbox{1}\quad\fbox{2}\quad\fbox{3}\quad\fbox{4}\quad\fbox{5}\quad\fbox{6}\quad\fbox{7}\quad\fbox{8}
\displaystyle \text{The }4\text{ vowels can be arranged in the }4\text{ odd positions in }4!=24\text{ ways.}
\displaystyle \text{The }4\text{ consonants can be arranged in the remaining }4\text{ positions in }4!=24\text{ ways.}
\displaystyle \therefore \text{Required number of arrangements}=4!\times4!=24\times24=576.
\displaystyle \\

\displaystyle \textbf{Question 5: } \text{How many words can be formed from the letters of the word}
\displaystyle \text{`SUNDAY'? How many of these begin with }N\text{? How many begin with }N
\displaystyle \text{and end with }Y\text{?}
\displaystyle \text{Answer:}
\displaystyle \text{The word }\text{SUNDAY}\text{ has }6\text{ distinct letters.}
\displaystyle \therefore \text{Total number of words that can be formed}={}^6P_6
\displaystyle =\frac{6!}{(6-6)!}=\frac{6!}{0!}=6!=720.
\displaystyle \text{If }N\text{ is fixed in the first position, the remaining }5\text{ letters can be arranged}
\displaystyle \text{in }{}^5P_5=\frac{5!}{(5-5)!}=\frac{5!}{0!}=5!=120\text{ ways.}
\displaystyle \text{If }N\text{ is fixed in the first position and }Y\text{ in the last position, the remaining}
\displaystyle 4\text{ letters can be arranged in }{}^4P_4=\frac{4!}{(4-4)!}=\frac{4!}{0!}=4!=24\text{ ways.}
\displaystyle \therefore \text{The required numbers are }720,\ 120\text{ and }24\text{ respectively.}
\displaystyle \\

\displaystyle \textbf{Question 6: } \text{How many different words can be formed from the letters of the word}
\displaystyle \text{`GANESHPURI'? In how many of these words:}
\displaystyle \text{(i) the letter }G\text{ always occupies the first position?}
\displaystyle \text{(ii) the letters }P\text{ and }I\text{ respectively occupy the first and last positions?}
\displaystyle \text{(iii) the vowels are always together?}
\displaystyle \text{(iv) the vowels always occupy even positions?}
\displaystyle \text{Answer:}
\displaystyle \text{The word }\text{GANESHPURI}\text{ has }10\text{ distinct letters.}
\displaystyle \therefore \text{Total number of words that can be formed}=10!=3628800.
\displaystyle \text{(i) If }G\text{ is fixed in the first position, the remaining }9\text{ letters can be arranged}
\displaystyle \text{in }9!=362880\text{ ways.}
\displaystyle \therefore \text{Number of words beginning with }G=362880.
\displaystyle \text{(ii) If }P\text{ is fixed in the first position and }I\text{ in the last position,}
\displaystyle \text{the remaining }8\text{ letters can be arranged in }8!=40320\text{ ways.}
\displaystyle \therefore \text{Required number of words}=40320.
\displaystyle \text{(iii) The vowels are }A,E,U,I\text{ and the consonants are }G,N,S,H,P,R.
\displaystyle \text{Treating the }4\text{ vowels as one block, there are }7\text{ objects to arrange.}
\displaystyle \text{These }7\text{ objects can be arranged in }7!\text{ ways.}
\displaystyle \text{The }4\text{ vowels within the block can be arranged in }4!\text{ ways.}
\displaystyle \therefore \text{Required number of arrangements}=7!\times4!
\displaystyle =5040\times24=120960.
\displaystyle \text{(iv) The five even positions are }2,4,6,8\text{ and }10.
\displaystyle \fbox{1}\quad\fbox{2}\quad\fbox{3}\quad\fbox{4}\quad\fbox{5}\quad\fbox{6}\quad\fbox{7}\quad\fbox{8}\quad\fbox{9}\quad\fbox{10}
\displaystyle \text{The }4\text{ vowels can be arranged in }4\text{ of the }5\text{ even positions in }{}^5P_4\text{ ways.}
\displaystyle {}^5P_4=\frac{5!}{(5-4)!}=\frac{5!}{1!}=120.
\displaystyle \text{The }6\text{ consonants can be arranged in the remaining }6\text{ positions in }6!=720\text{ ways.}
\displaystyle \therefore \text{Required number of arrangements}=120\times720=86400.
\displaystyle \\

\displaystyle \textbf{Question 7: } \text{How many permutations can be formed using the letters of the word}
\displaystyle \text{`VOWELS' when:}
\displaystyle \text{i) there is no restriction on the letters?}
\displaystyle \text{ii) each word begins with the letter }E\text{?}
\displaystyle \text{iii) each word begins with }O\text{ and ends with }L\text{?}
\displaystyle \text{iv) all the vowels come together?}
\displaystyle \text{v) all the consonants come together?}
\displaystyle \text{Answer:}
\displaystyle \text{The word }\text{VOWELS}\text{ has }6\text{ distinct letters.}
\displaystyle \text{i) When there is no restriction, the }6\text{ letters can be arranged in }6!\text{ ways.}
\displaystyle \therefore \text{Required number of permutations}={}^6P_6=\frac{6!}{0!}=6!=720.
\displaystyle \text{ii) If }E\text{ is fixed in the first position, the remaining }5\text{ letters can be arranged}
\displaystyle \text{in }{}^5P_5=\frac{5!}{0!}=5!=120\text{ ways.}
\displaystyle \therefore \text{Required number of permutations}=120.
\displaystyle \text{iii) If }O\text{ is fixed in the first position and }L\text{ in the last position,}
\displaystyle \text{the remaining }4\text{ letters can be arranged in }{}^4P_4=\frac{4!}{0!}=4!=24\text{ ways.}
\displaystyle \therefore \text{Required number of permutations}=24.
\displaystyle \text{iv) The vowels are }O,E\text{ and the consonants are }V,W,L,S.
\displaystyle \text{Treating the }2\text{ vowels as one block, there are }5\text{ objects to arrange.}
\displaystyle \text{These }5\text{ objects can be arranged in }5!\text{ ways.}
\displaystyle \text{The }2\text{ vowels within the block can be arranged in }2!\text{ ways.}
\displaystyle \therefore \text{Required number of permutations}=5!\times2!=120\times2=240.
\displaystyle \text{v) Treating the }4\text{ consonants as one block, there are }3\text{ objects to arrange.}
\displaystyle \text{These }3\text{ objects can be arranged in }3!\text{ ways.}
\displaystyle \text{The }4\text{ consonants within the block can be arranged in }4!\text{ ways.}
\displaystyle \therefore \text{Required number of permutations}=3!\times4!=6\times24=144.
\displaystyle \\

\displaystyle \textbf{Question 8: } \text{How many words can be formed from the letters of the word}
\displaystyle \text{`ARTICLE' so that the vowels occupy the even positions?}
\displaystyle \text{Answer:}
\displaystyle \text{The word }\text{ARTICLE}\text{ has }7\text{ distinct letters.}
\displaystyle \text{The vowels are }A,I,E\text{ and the consonants are }R,T,C,L.
\displaystyle \fbox{1}\quad\fbox{2}\quad\fbox{3}\quad\fbox{4}\quad\fbox{5}\quad\fbox{6}\quad\fbox{7}
\displaystyle \text{The three even positions are }2,4\text{ and }6.
\displaystyle \text{The }3\text{ vowels can be arranged in the }3\text{ even positions in }3!\text{ ways.}
\displaystyle {}^3P_3=\frac{3!}{(3-3)!}=\frac{3!}{0!}=3!=6.
\displaystyle \text{The }4\text{ consonants can be arranged in the remaining }4\text{ positions in }4!=24\text{ ways.}
\displaystyle \therefore \text{Required number of arrangements}=3!\times4!=6\times24=144.
\displaystyle \\

\displaystyle \textbf{Question 9: } \text{In how many ways can a game of lawn tennis mixed doubles be}
\displaystyle \text{formed from seven married couples if no husband and wife play in the same game?}
\displaystyle \text{Answer:}
\displaystyle \text{Choose }2\text{ husbands from the }7\text{ husbands in }{}^7C_2\text{ ways.}
\displaystyle \text{The wives of the selected husbands cannot be chosen.}
\displaystyle \text{Therefore, choose }2\text{ wives from the remaining }5\text{ wives in }{}^5C_2\text{ ways.}
\displaystyle \text{The selected }2\text{ husbands and }2\text{ wives can be paired into two mixed teams in }2!\text{ ways.}
\displaystyle \therefore \text{Required number of games}={}^7C_2\times{}^5C_2\times2!
\displaystyle =21\times10\times2=420.
\displaystyle \\

\displaystyle \textbf{Question 10: } m\text{ men and }n\text{ women are to be seated in a row so that no two}
\displaystyle \text{women sit together. If }m>n,\text{ show that the number of ways in which they}
\displaystyle \text{can be seated is }\frac{m!(m+1)!}{(m-n+1)!}.
\displaystyle \text{Answer:}
\displaystyle \text{The }m\text{ men can be arranged in a row in }m!\text{ ways.}
\displaystyle \text{After arranging the men, there are }m+1\text{ gaps in which the women can sit.}
\displaystyle \_\,M\,\_\,M\,\_\,\ldots\,M\,\_
\displaystyle \text{To ensure that no two women sit together, each woman must occupy a different gap.}
\displaystyle \text{The }n\text{ women can be arranged in }n\text{ of the }m+1\text{ gaps in }{}^{m+1}P_n\text{ ways.}
\displaystyle \therefore \text{Required number of arrangements}=m!\times{}^{m+1}P_n
\displaystyle =m!\times\frac{(m+1)!}{(m+1-n)!}
\displaystyle =\frac{m!(m+1)!}{(m-n+1)!}.
\displaystyle \therefore \text{The given result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 11: } \text{How many words, with or without dictionary meaning, can be made}
\displaystyle \text{from the letters of the word }\text{`MONDAY', assuming that no letter is repeated, if:}
\displaystyle \text{(i) }4\text{ letters are used at a time?}
\displaystyle \text{(ii) all the letters are used at a time?}
\displaystyle \text{(iii) all the letters are used and the first letter is a vowel?}
\displaystyle \text{Answer:}
\displaystyle \text{The word }\text{MONDAY}\text{ has }6\text{ distinct letters.}
\displaystyle \text{(i) Number of }4\text{-letter words}={}^6P_4
\displaystyle =\frac{6!}{(6-4)!}=\frac{6!}{2!}=6\times5\times4\times3=360.
\displaystyle \text{(ii) Number of }6\text{-letter words}={}^6P_6
\displaystyle =\frac{6!}{(6-6)!}=\frac{6!}{0!}=6!=720.
\displaystyle \text{(iii) The vowels in }\text{MONDAY}\text{ are }O\text{ and }A.
\displaystyle \text{Therefore, there are }2\text{ choices for the first position.}
\displaystyle \text{The remaining }5\text{ letters can be arranged in }5!\text{ ways.}
\displaystyle \therefore \text{Required number of words}=2\times5!=2\times120=240.
\displaystyle \\

\displaystyle \textbf{Question 12: } \text{How many three-letter words can be made using the letters of the}
\displaystyle \text{word }\text{`ORIENTAL'?}
\displaystyle \text{Answer:}
\displaystyle \text{The word }\text{ORIENTAL}\text{ has }8\text{ distinct letters.}
\displaystyle \text{The number of three-letter words that can be formed}={}^8P_3
\displaystyle =\frac{8!}{(8-3)!}=\frac{8!}{5!}=8\times7\times6=336.
\displaystyle \\

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