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

\displaystyle \text{i) INDEPENDENCE}
\displaystyle \text{The word INDEPENDENCE has }12\text{ letters, of which E occurs }4\text{ times,}
\displaystyle \text{N occurs }3\text{ times and D occurs }2\text{ times.}
\displaystyle \therefore \text{Number of words}=\frac{12!}{4!3!2!}
\displaystyle =\frac{479001600}{24\times6\times2}=1663200

\displaystyle \text{ii) INTERMEDIATE}
\displaystyle \text{The word INTERMEDIATE has }12\text{ letters, of which E occurs }3\text{ times,}
\displaystyle \text{I occurs }2\text{ times and T occurs }2\text{ times.}
\displaystyle \therefore \text{Number of words}=\frac{12!}{3!2!2!}
\displaystyle =\frac{479001600}{6\times2\times2}=19958400

\displaystyle \text{iii) ARRANGE}
\displaystyle \text{The word ARRANGE has }7\text{ letters, of which A occurs }2\text{ times and}
\displaystyle \text{R occurs }2\text{ times.}
\displaystyle \therefore \text{Number of words}=\frac{7!}{2!2!}
\displaystyle =\frac{5040}{2\times2}=1260

\displaystyle \text{iv) INDIA}
\displaystyle \text{The word INDIA has }5\text{ letters, of which I occurs }2\text{ times.}
\displaystyle \therefore \text{Number of words}=\frac{5!}{2!}
\displaystyle =\frac{120}{2}=60

\displaystyle \text{v) PAKISTAN}
\displaystyle \text{The word PAKISTAN has }8\text{ letters, of which A occurs }2\text{ times.}
\displaystyle \therefore \text{Number of words}=\frac{8!}{2!}
\displaystyle =\frac{40320}{2}=20160

\displaystyle \text{vi) RUSSIA}
\displaystyle \text{The word RUSSIA has }6\text{ letters, of which S occurs }2\text{ times.}
\displaystyle \therefore \text{Number of words}=\frac{6!}{2!}
\displaystyle =\frac{720}{2}=360

\displaystyle \text{vii) SERIES}
\displaystyle \text{The word SERIES has }6\text{ letters, of which S occurs }2\text{ times and}
\displaystyle \text{E occurs }2\text{ times.}
\displaystyle \therefore \text{Number of words}=\frac{6!}{2!2!}
\displaystyle =\frac{720}{2\times2}=180

\displaystyle \text{viii) EXERCISES}
\displaystyle \text{The word EXERCISES has }9\text{ letters, of which E occurs }3\text{ times and}
\displaystyle \text{S occurs }2\text{ times.}
\displaystyle \therefore \text{Number of words}=\frac{9!}{3!2!}
\displaystyle =\frac{362880}{6\times2}=30240

\displaystyle \text{ix) CONSTANTINOPLE}
\displaystyle \text{The word CONSTANTINOPLE has }14\text{ letters, of which N occurs }3\text{ times,}
\displaystyle \text{O occurs }2\text{ times and T occurs }2\text{ times.}
\displaystyle \therefore \text{Number of words}=\frac{14!}{3!2!2!}
\displaystyle =\frac{87178291200}{6\times2\times2}=3632428800
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{In how many ways can the letters of the word ALGEBRA be arranged without}
\displaystyle \text{changing the relative order of the vowels and consonants?}
\displaystyle \text{Answer:}
\displaystyle \text{In the word ALGEBRA, the vowels occur in the relative order A, E, A,}
\displaystyle \text{and the consonants occur in the relative order L, G, B, R.}
\displaystyle \text{Since their relative orders cannot be changed, we only need to choose the}
\displaystyle \text{three positions occupied by the vowels out of the seven available positions.}
\displaystyle \therefore \text{Number of arrangements}={}^{7}C_{3}
\displaystyle =\frac{7!}{3!4!}
\displaystyle =\frac{7\times6\times5}{3\times2\times1}=35
\displaystyle \therefore \text{The required number of arrangements is }35.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{How many words can be formed with the letters of the word UNIVERSITY,}
\displaystyle \text{if all the vowels remain together?}
\displaystyle \text{Answer:}
\displaystyle \text{The word UNIVERSITY has }10\text{ letters, of which the vowels are U, I, E, I.}
\displaystyle \text{There are }4\text{ vowels, of which I occurs twice.}
\displaystyle \text{Treating all the vowels as one block, we have one vowel block and}
\displaystyle 6\text{ consonants, giving a total of }7\text{ objects.}
\displaystyle \text{Number of ways to arrange these }7\text{ objects}=7!
\displaystyle \text{Number of ways to arrange the vowels within the block}=\frac{4!}{2!}=12
\displaystyle \therefore \text{Required number of words}=7!\times\frac{4!}{2!}
\displaystyle =5040\times12=60480
\displaystyle \therefore \text{The required number of words is }60480.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the total number of arrangements of the letters in the expression}
\displaystyle a^3b^2c^4\text{ when written at full length.}
\displaystyle \text{Answer:}
\displaystyle \text{When written at full length, the expression is }aaabbcccc.
\displaystyle \text{It contains }9\text{ letters, of which a occurs }3\text{ times, b occurs }2\text{ times}
\displaystyle \text{and c occurs }4\text{ times.}
\displaystyle \therefore \text{Number of distinct arrangements}=\frac{9!}{3!2!4!}
\displaystyle =\frac{9\times8\times7\times6\times5\times4!}{3!\times2!\times4!}
\displaystyle =\frac{9\times8\times7\times6\times5}{6\times2}
\displaystyle =1260
\displaystyle \therefore \text{The total number of arrangements is }1260.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{How many words can be formed with the letters of the word PARALLEL}
\displaystyle \text{such that all the L's do not come together?}
\displaystyle \text{Answer:}
\displaystyle \text{The word PARALLEL has }8\text{ letters, of which A occurs }2\text{ times and}
\displaystyle \text{L occurs }3\text{ times.}
\displaystyle \text{Total number of distinct words}=\frac{8!}{2!3!}
\displaystyle =\frac{40320}{2\times6}=3360
\displaystyle \text{Now, consider the arrangements in which all three L's are together.}
\displaystyle \text{Treating LLL as one block, we have }6\text{ objects, of which A occurs twice.}
\displaystyle \therefore \text{Number of words in which all the L's are together}=\frac{6!}{2!}
\displaystyle =\frac{720}{2}=360
\displaystyle \therefore \text{Required number of words}=3360-360
\displaystyle =3000
\displaystyle \therefore \text{The required number of words is }3000.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{How many words can be formed by arranging the letters of the word}
\displaystyle \text{MUMBAI so that both the M's come together?}
\displaystyle \text{Answer:}
\displaystyle \text{The word MUMBAI has }6\text{ letters, of which M occurs twice.}
\displaystyle \text{Treating both the M's as one block, we have }5\text{ distinct objects.}
\displaystyle \therefore \text{Number of arrangements}=5!
\displaystyle =5\times4\times3\times2\times1=120
\displaystyle \therefore \text{The required number of words is }120.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{How many numbers can be formed with the digits }1,2,3,4,3,2,1
\displaystyle \text{such that the odd digits always occupy the odd places?}
\displaystyle \text{Answer:}
\displaystyle \text{There are }7\text{ digits and hence }4\text{ odd places, namely the first, third, fifth}
\displaystyle \text{and seventh places.}
\displaystyle \text{The odd digits are }1,1,3,3.
\displaystyle \text{Number of ways to arrange the odd digits in the odd places}=\frac{4!}{2!2!}
\displaystyle =\frac{24}{4}=6
\displaystyle \text{The even digits are }2,2,4.
\displaystyle \text{Number of ways to arrange the even digits in the even places}=\frac{3!}{2!}
\displaystyle =\frac{6}{2}=3
\displaystyle \therefore \text{Required number of numbers}=6\times3=18
\displaystyle \therefore \text{The required number of numbers is }18.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{How many different signals can be made from }4\text{ red, }2\text{ white and}
\displaystyle 3\text{ green flags by arranging all of them vertically on a flagstaff?}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of flags}=4+2+3=9
\displaystyle \text{There are }4\text{ identical red flags, }2\text{ identical white flags and}
\displaystyle 3\text{ identical green flags.}
\displaystyle \therefore \text{Number of different signals}=\frac{9!}{4!2!3!}
\displaystyle =\frac{9\times8\times7\times6\times5\times4!}{4!\times2!\times3!}
\displaystyle =\frac{9\times8\times7\times6\times5}{2\times6}
\displaystyle =1260
\displaystyle \therefore \text{The required number of signals is }1260.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{How many four-digit numbers can be formed using the digits }1,3,3,0?
\displaystyle \text{Answer:}
\displaystyle \text{The digits are }1,3,3,0,\text{ where the digit }3\text{ occurs twice.}
\displaystyle \text{Total number of distinct arrangements}=\frac{4!}{2!}
\displaystyle =\frac{24}{2}=12
\displaystyle \text{A four-digit number cannot begin with }0.
\displaystyle \text{If }0\text{ is fixed in the first place, the remaining digits }1,3,3\text{ can be arranged in}
\displaystyle \frac{3!}{2!}=\frac{6}{2}=3\text{ ways.}
\displaystyle \therefore \text{Required number of four-digit numbers}=12-3
\displaystyle =9
\displaystyle \therefore \text{The required number of four-digit numbers is }9.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{In how many ways can the letters of the word ARRANGE be arranged so that}
\displaystyle \text{the two R's are never together?}
\displaystyle \text{Answer:}
\displaystyle \text{The word ARRANGE has }7\text{ letters, of which A occurs twice and R occurs twice.}
\displaystyle \text{Total number of distinct arrangements}=\frac{7!}{2!2!}
\displaystyle =\frac{5040}{2\times2}=1260
\displaystyle \text{Now, consider the arrangements in which the two R's are together.}
\displaystyle \text{Treating RR as one block, we have }6\text{ objects, of which A occurs twice.}
\displaystyle \therefore \text{Number of arrangements in which the two R's are together}=\frac{6!}{2!}
\displaystyle =\frac{720}{2}=360
\displaystyle \therefore \text{Number of arrangements in which the two R's are never together}
\displaystyle =1260-360=900
\displaystyle \therefore \text{The required number of arrangements is }900.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{How many different numbers greater than }50000\text{ can be formed using}
\displaystyle \text{all the digits }0,1,1,5,9?
\displaystyle \text{Answer:}
\displaystyle \text{For the number to be greater than }50000,\text{ the first digit must be either }5\text{ or }9.
\displaystyle \text{When }5\text{ is in the first place, the remaining digits }0,1,1,9\text{ can be arranged in}
\displaystyle \frac{4!}{2!}=\frac{24}{2}=12\text{ ways.}
\displaystyle \text{When }9\text{ is in the first place, the remaining digits }0,1,1,5\text{ can be arranged in}
\displaystyle \frac{4!}{2!}=\frac{24}{2}=12\text{ ways.}
\displaystyle \therefore \text{Required number of different numbers}=12+12
\displaystyle =24
\displaystyle \therefore \text{The required number of numbers is }24.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{How many words can be formed from the letters of the word SERIES which}
\displaystyle \text{start with S and end with S?}
\displaystyle \text{Answer:}
\displaystyle \text{The word SERIES contains two S's and two E's.}
\displaystyle \text{Fixing one S at the beginning and the other S at the end, we get}
\displaystyle \fbox{S}\quad\fbox{\phantom{E}}\quad\fbox{\phantom{R}}\quad\fbox{\phantom{I}}\quad\fbox{\phantom{E}}\quad\fbox{S}
\displaystyle \text{The four middle places are to be filled with E, E, R and I.}
\displaystyle \therefore \text{Number of words}=\frac{4!}{2!}
\displaystyle =\frac{24}{2}=12
\displaystyle \therefore \text{The required number of words is }12.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{How many permutations of the letters of the word MADHUBANI do not}
\displaystyle \text{begin with M but end with I?}
\displaystyle \text{Answer:}
\displaystyle \text{The word MADHUBANI has }9\text{ letters, of which A occurs twice.}
\displaystyle \text{Fixing I at the end, the remaining }8\text{ letters can be arranged in}
\displaystyle \frac{8!}{2!}=\frac{40320}{2}=20160\text{ ways.}
\displaystyle \text{Now, fix M at the beginning and I at the end.}
\displaystyle \text{The remaining }7\text{ letters can be arranged in}
\displaystyle \frac{7!}{2!}=\frac{5040}{2}=2520\text{ ways.}
\displaystyle \therefore \text{Number of permutations which do not begin with M but end with I}
\displaystyle =20160-2520
\displaystyle =17640
\displaystyle \therefore \text{The required number of permutations is }17640.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Find the number of numbers greater than one million that can be formed}
\displaystyle \text{using all the digits }2,3,0,3,4,2,3.
\displaystyle \text{Answer:}
\displaystyle \text{There are }7\text{ digits, of which }2\text{ occurs twice and }3\text{ occurs three times.}
\displaystyle \text{Total number of distinct arrangements of all the digits}=\frac{7!}{2!3!}
\displaystyle =\frac{5040}{2\times6}=420
\displaystyle \text{However, an arrangement beginning with }0\text{ does not form a seven-digit number.}
\displaystyle \text{Fixing }0\text{ in the first place, the remaining digits can be arranged in}
\displaystyle \frac{6!}{2!3!}=\frac{720}{2\times6}=60\text{ ways.}
\displaystyle \therefore \text{Required number of numbers}=420-60
\displaystyle =360
\displaystyle \therefore \text{The required number of numbers is }360.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{There are three copies each of }4\text{ different books. In how many ways}
\displaystyle \text{can they be arranged on a shelf?}
\displaystyle \text{Answer:}
\displaystyle \text{There are three identical copies of each of }4\text{ different books.}
\displaystyle \therefore \text{Total number of books}=3\times4=12
\displaystyle \text{Each of the four different books occurs three times.}
\displaystyle \therefore \text{Number of distinct arrangements}=\frac{12!}{3!3!3!3!}
\displaystyle =\frac{479001600}{6\times6\times6\times6}
\displaystyle =369600
\displaystyle \therefore \text{The required number of arrangements is }369600.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{How many different arrangements can be made using all the letters of}
\displaystyle \text{the word MATHEMATICS? How many of them begin with C? How many of them}
\displaystyle \text{begin with T?}
\displaystyle \text{Answer:}
\displaystyle \text{The word MATHEMATICS has }11\text{ letters, of which M, A and T occur twice each,}
\displaystyle \text{while the remaining letters are distinct.}
\displaystyle \therefore \text{Total number of different arrangements}=\frac{11!}{2!2!2!}
\displaystyle =\frac{39916800}{2\times2\times2}=4989600
\displaystyle \text{To find the number of arrangements beginning with C, fix C in the first place.}
\displaystyle \text{The remaining }10\text{ letters contain two M's, two A's and two T's.}
\displaystyle \therefore \text{Number of arrangements beginning with C}=\frac{10!}{2!2!2!}
\displaystyle =\frac{3628800}{2\times2\times2}=453600
\displaystyle \text{To find the number of arrangements beginning with T, fix one T in the first place.}
\displaystyle \text{The remaining }10\text{ letters contain two M's and two A's.}
\displaystyle \therefore \text{Number of arrangements beginning with T}=\frac{10!}{2!2!}
\displaystyle =\frac{3628800}{2\times2}=907200
\displaystyle \therefore \text{The total number of arrangements is }4989600.
\displaystyle \therefore \text{The number beginning with C is }453600\text{ and the number beginning with T is }907200.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{A biologist studying the genetic code is interested in the number of possible}
\displaystyle \text{arrangements of }12\text{ molecules in a chain. The chain contains four different molecules}
\displaystyle \text{represented by A, C, G and T, with three molecules of each kind. How many}
\displaystyle \text{different arrangements are possible?}
\displaystyle \text{Answer:}
\displaystyle \text{The chain contains }12\text{ molecules, with three molecules each of A, C, G and T.}
\displaystyle \therefore \text{Number of different arrangements}=\frac{12!}{3!3!3!3!}
\displaystyle =\frac{479001600}{6\times6\times6\times6}
\displaystyle =369600
\displaystyle \therefore \text{The required number of arrangements is }369600.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{In how many ways can }4\text{ red, }3\text{ yellow and }2\text{ green discs be}
\displaystyle \text{arranged in a row if the discs of the same colour are indistinguishable?}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of discs}=4+3+2=9
\displaystyle \text{There are }4\text{ identical red discs, }3\text{ identical yellow discs and}
\displaystyle 2\text{ identical green discs.}
\displaystyle \therefore \text{Number of distinct arrangements}=\frac{9!}{4!3!2!}
\displaystyle =\frac{9\times8\times7\times6\times5\times4!}{4!\times3!\times2!}
\displaystyle =\frac{9\times8\times7\times6\times5}{6\times2}
\displaystyle =1260
\displaystyle \therefore \text{The required number of arrangements is }1260.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{How many numbers greater than }1000000\text{ can be formed using all the}
\displaystyle \text{digits }1,2,0,2,4,2,4?
\displaystyle \text{Answer:}
\displaystyle \text{There are }7\text{ digits, of which }2\text{ occurs three times and }4\text{ occurs twice.}
\displaystyle \text{Total number of distinct arrangements of all the digits}=\frac{7!}{3!2!}
\displaystyle =\frac{5040}{6\times2}=420
\displaystyle \text{An arrangement beginning with }0\text{ is not a seven-digit number.}
\displaystyle \text{Fixing }0\text{ in the first place, the remaining digits can be arranged in}
\displaystyle \frac{6!}{3!2!}=\frac{720}{6\times2}=60\text{ ways.}
\displaystyle \therefore \text{Required number of numbers}=420-60
\displaystyle =360
\displaystyle \therefore \text{The required number of numbers is }360.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{In how many ways can the letters of the word ASSASSINATION be arranged}
\displaystyle \text{so that all the S's are together?}
\displaystyle \text{Answer:}
\displaystyle \text{The word ASSASSINATION has }13\text{ letters, of which A occurs }3\text{ times,}
\displaystyle \text{S occurs }4\text{ times, I occurs }2\text{ times and N occurs }2\text{ times.}
\displaystyle \text{Treating all four S's as one block, we have }10\text{ objects to arrange.}
\displaystyle \text{Among these objects, A occurs }3\text{ times, I occurs }2\text{ times and N occurs }2\text{ times.}
\displaystyle \therefore \text{Number of arrangements}=\frac{10!}{3!2!2!}
\displaystyle =\frac{3628800}{6\times2\times2}
\displaystyle =151200
\displaystyle \therefore \text{The required number of arrangements is }151200.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Find the total number of permutations of the letters of the word INSTITUTE.}
\displaystyle \text{Answer:}
\displaystyle \text{The word INSTITUTE has }9\text{ letters, of which I occurs twice and T occurs}
\displaystyle 3\text{ times, while the remaining letters are distinct.}
\displaystyle \therefore \text{Total number of permutations}=\frac{9!}{2!3!}
\displaystyle =\frac{362880}{2\times6}
\displaystyle =30240
\displaystyle \therefore \text{The total number of permutations is }30240.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{The letters of the word SURITI are written in all possible orders and}
\displaystyle \text{these words are arranged as in a dictionary. Find the rank of the word SURITI.}
\displaystyle \text{Answer:}
\displaystyle \text{The distinct letters are }I,\;R,\;S,\;T,\;U,\text{ with I occurring twice.}
\displaystyle \text{The alphabetical order is }I,\;R,\;S,\;T,\;U.
\displaystyle \text{Words beginning with I}=\frac{5!}{1!}=120
\displaystyle \text{Words beginning with R}=\frac{5!}{2!}=60
\displaystyle \text{Now consider words beginning with S.}
\displaystyle \text{Words beginning with SI}=4!=24
\displaystyle \text{Words beginning with SR}=\frac{4!}{2!}=12
\displaystyle \text{Words beginning with ST}=\frac{4!}{2!}=12
\displaystyle \text{Now consider words beginning with SU.}
\displaystyle \text{Words beginning with SUI}=3!=6
\displaystyle \text{Now consider words beginning with SURI.}
\displaystyle \text{At the fifth position, the word SURIIT comes before SURITI.}
\displaystyle \therefore \text{Number of words before SURITI}=120+60+24+12+12+6+1=235
\displaystyle \therefore \text{Rank of the word SURITI}=235+1=236
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{If the letters of the word LATE are permuted and the words so formed}
\displaystyle \text{are arranged as in a dictionary, find the rank of the word LATE.}
\displaystyle \text{Answer:}
\displaystyle \text{The letters in alphabetical order are }A,\;E,\;L,\;T.
\displaystyle \text{Words beginning with A}=3!=6
\displaystyle \text{Words beginning with E}=3!=6
\displaystyle \text{Now consider words beginning with L.}
\displaystyle \text{The second letter of LATE is A, so no word beginning with LE comes before it.}
\displaystyle \text{Fixing LA, the remaining letters are E and T.}
\displaystyle \text{The only word before LATE is LAET.}
\displaystyle \therefore \text{Number of words before LATE}=6+6+1=13
\displaystyle \therefore \text{Rank of the word LATE}=13+1=14
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{If the letters of the word MOTHER are written in all possible orders}
\displaystyle \text{and these words are arranged as in a dictionary, find the rank of MOTHER.}
\displaystyle \text{Answer:}
\displaystyle \text{The letters in alphabetical order are }E,\;H,\;M,\;O,\;R,\;T.
\displaystyle \text{Words beginning with E}=5!=120
\displaystyle \text{Words beginning with H}=5!=120
\displaystyle \text{Now consider words beginning with M.}
\displaystyle \text{The second letter of MOTHER is O.}
\displaystyle \text{Words beginning with ME}=4!=24
\displaystyle \text{Words beginning with MH}=4!=24
\displaystyle \text{Now consider words beginning with MO.}
\displaystyle \text{The third letter of MOTHER is T.}
\displaystyle \text{Words beginning with MOE}=3!=6
\displaystyle \text{Words beginning with MOH}=3!=6
\displaystyle \text{Words beginning with MOR}=3!=6
\displaystyle \text{Now consider words beginning with MOT.}
\displaystyle \text{The fourth letter of MOTHER is H. Of the remaining letters E, H and R,}
\displaystyle \text{only E comes before H.}
\displaystyle \text{Words beginning with MOTE}=2!=2
\displaystyle \therefore \text{Number of words before MOTHER}
\displaystyle =120+120+24+24+6+6+6+2
\displaystyle =308
\displaystyle \therefore \text{Rank of the word MOTHER}=308+1=309
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{If the permutations of }a,b,c,d,e\text{ taken all together are written}
\displaystyle \text{in alphabetical order as in a dictionary and numbered, find the rank of debac.}
\displaystyle \text{Answer:}
\displaystyle \text{The letters in alphabetical order are }a,\;b,\;c,\;d,\;e.
\displaystyle \text{Words beginning with a}=4!=24
\displaystyle \text{Words beginning with b}=4!=24
\displaystyle \text{Words beginning with c}=4!=24
\displaystyle \text{Now consider words beginning with d.}
\displaystyle \text{Words beginning with da}=3!=6
\displaystyle \text{Words beginning with db}=3!=6
\displaystyle \text{Words beginning with dc}=3!=6
\displaystyle \text{Now consider words beginning with de.}
\displaystyle \text{The permutations beginning with de up to debac are deabc, deacb and debac.}
\displaystyle \therefore \text{Rank of debac}=24+24+24+6+6+6+3
\displaystyle =93
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{Find the total number of ways in which six }+\text{ signs and}
\displaystyle \text{four }-\text{ signs} \ \text{can be arranged in a line such that no two }-\text{ signs occur together.}
\displaystyle \text{Answer:}
\displaystyle \text{Arrange the six identical }+\text{ signs in a row.}
\displaystyle +\quad+\quad+\quad+\quad+\quad+
\displaystyle \text{There are }7\text{ gaps in which the }-\text{ signs can be placed.}
\displaystyle \text{To ensure that no two }-\text{ signs are together, choose }4\text{ of these }7\text{ gaps.}
\displaystyle \therefore \text{Number of arrangements}={}^{7}C_4
\displaystyle =\frac{7!}{4!3!}
\displaystyle =35
\displaystyle \therefore \text{The required number of arrangements is }35.
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{In how many ways can the letters of the word INTERMEDIATE be arranged}
\displaystyle \text{so that:}
\displaystyle \text{(i) the vowels always occupy the even places?}
\displaystyle \text{(ii) the relative order of the vowels and consonants does not change?}
\displaystyle \text{Answer:}
\displaystyle \text{The word INTERMEDIATE has }12\text{ letters.}
\displaystyle \text{The vowels are I, E, E, I, A, E and the consonants are N, T, R, M, D, T.}
\displaystyle \text{Thus, there are }6\text{ vowels and }6\text{ consonants.}
\displaystyle \text{(i) The six vowels must occupy the six even places.}
\displaystyle \text{Number of ways to arrange the vowels}=\frac{6!}{2!3!}
\displaystyle =\frac{720}{2\times6}=60
\displaystyle \text{Number of ways to arrange the consonants}=\frac{6!}{2!}
\displaystyle =\frac{720}{2}=360
\displaystyle \therefore \text{Required number of arrangements}=60\times360
\displaystyle =21600
\displaystyle \text{(ii) The relative order of the vowels and consonants must remain unchanged.}
\displaystyle \text{Therefore, we only need to choose }6\text{ of the }12\text{ places for the vowels.}
\displaystyle \text{The consonants will occupy the remaining }6\text{ places in their original order.}
\displaystyle \therefore \text{Required number of arrangements}={}^{12}C_6
\displaystyle =\frac{12!}{6!6!}
\displaystyle =924
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{The letters of the word ZENITH are written in all possible orders. How many}
\displaystyle \text{words are possible? If these words are arranged as in a dictionary, find the rank}
\displaystyle \text{of the word ZENITH.}
\displaystyle \text{Answer:}
\displaystyle \text{The word ZENITH has }6\text{ distinct letters.}
\displaystyle \therefore \text{Total number of possible words}=6!=720
\displaystyle \text{The letters in alphabetical order are }E,\;H,\;I,\;N,\;T,\;Z.
\displaystyle \text{Words beginning with E, H, I, N or T}=5\times5!
\displaystyle =5\times120=600
\displaystyle \text{Now consider words beginning with Z.}
\displaystyle \text{The second letter of ZENITH is E, the smallest remaining letter.}
\displaystyle \text{Therefore, no additional word precedes ZENITH at the second position.}
\displaystyle \text{Now consider words beginning with ZE.}
\displaystyle \text{The third letter is N. Of the remaining letters H, I, N and T,}
\displaystyle \text{the letters H and I come before N.}
\displaystyle \text{Number of such words}=2\times3!=12
\displaystyle \text{Now consider words beginning with ZEN.}
\displaystyle \text{The fourth letter is I. Of the remaining letters H, I and T,}
\displaystyle \text{only H comes before I.}
\displaystyle \text{Number of such words}=2!=2
\displaystyle \text{Now consider words beginning with ZENI.}
\displaystyle \text{The fifth letter is T. Of the remaining letters H and T,}
\displaystyle \text{only H comes before T.}
\displaystyle \text{Number of such words}=1
\displaystyle \therefore \text{Number of words before ZENITH}=600+12+2+1
\displaystyle =615
\displaystyle \therefore \text{Rank of the word ZENITH}=615+1=616
\displaystyle \\

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