\displaystyle \textbf{Question 1: }\text{How many different words, each containing }2\text{ vowels and }3\text{ consonants,}
\displaystyle \text{can be formed from }5\text{ vowels and }17\text{ consonants?}
\displaystyle \text{Answer:}
\displaystyle \text{The }2\text{ vowels can be chosen from }5\text{ vowels in }{}^{5}\mathrm{C}_{2}\text{ ways.}
\displaystyle \text{The }3\text{ consonants can be chosen from }17\text{ consonants in }{}^{17}\mathrm{C}_{3}\text{ ways.}
\displaystyle \text{The }5\text{ selected letters can be arranged in }5!\text{ ways.}
\displaystyle \therefore \text{Required number of words}
\displaystyle ={}^{5}\mathrm{C}_{2}\times{}^{17}\mathrm{C}_{3}\times5!
\displaystyle =\frac{5!}{2!\,3!}\times\frac{17!}{3!\,14!}\times5!
\displaystyle =10\times680\times120
\displaystyle =6800\times120
\displaystyle =816000
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{There are }10\text{ persons named }P_1,P_2,P_3,\ldots,P_{10}.
\displaystyle \text{Out of these, }5\text{ persons are to be arranged in a line such that }P_1\text{ must}
\displaystyle \text{occur, whereas }P_4\text{ and }P_5\text{ do not occur. Find the number of such arrangements.}
\displaystyle \text{Answer:}
\displaystyle P_1\text{ must be included, while }P_4\text{ and }P_5\text{ must be excluded.}
\displaystyle \therefore \text{The remaining }4\text{ persons are selected from the other }7\text{ persons.}
\displaystyle \text{Number of ways to select these }4\text{ persons}={}^{7}\mathrm{C}_{4}
\displaystyle =\frac{7!}{4!\,3!}
\displaystyle =\frac{7\times6\times5}{3\times2\times1}
\displaystyle =35
\displaystyle \text{The selected }5\text{ persons can be arranged in a line in }5!\text{ ways.}
\displaystyle \therefore \text{Required number of arrangements}={}^{7}\mathrm{C}_{4}\times5!
\displaystyle =35\times120
\displaystyle =4200
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{How many words, with or without meaning, can be formed from the letters}
\displaystyle \text{of the word MONDAY, assuming that no letter is repeated, if (i) }4\text{ letters are}
\displaystyle \text{used at a time; (ii) all letters are used; (iii) all letters are used but the first}
\displaystyle \text{letter is a vowel?}
\displaystyle \text{Answer:}
\displaystyle \text{i) Select }4\text{ letters from }6\text{ letters and arrange them.}
\displaystyle \therefore \text{Required number of words}={}^{6}\mathrm{C}_{4}\times4!
\displaystyle =15\times24
\displaystyle =360
\displaystyle \text{ii) If all }6\text{ letters are used, they can be arranged in}
\displaystyle 6!=720\text{ ways.}
\displaystyle \text{iii) The first letter must be a vowel.}
\displaystyle \text{The vowels are A and O, so the first position can be filled in }2\text{ ways.}
\displaystyle \text{The remaining }5\text{ letters can be arranged in }5!\text{ ways.}
\displaystyle \therefore \text{Required number of words}=2\times5!
\displaystyle =2\times120
\displaystyle =240
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the number of permutations of }n\text{ distinct things taken }r\text{ at a time,}
\displaystyle \text{in which }3\text{ particular things must occur together.}
\displaystyle \text{Answer:}
\displaystyle \text{The }3\text{ particular things must be included in every selection.}
\displaystyle \therefore \text{The remaining }r-3\text{ things are selected from the other }n-3\text{ things.}
\displaystyle \text{Number of ways to select them}={}^{n-3}\mathrm{C}_{r-3}
\displaystyle \text{Treat the }3\text{ particular things as one unit.}
\displaystyle \text{Thus, the }r-3\text{ selected things and the one unit form }r-2\text{ objects.}
\displaystyle \text{These }r-2\text{ objects can be arranged in }(r-2)!\text{ ways.}
\displaystyle \text{The }3\text{ particular things can be arranged among themselves in }3!\text{ ways.}
\displaystyle \therefore \text{Required number of permutations}
\displaystyle ={}^{n-3}\mathrm{C}_{r-3}\times(r-2)!\times3!
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{How many words, each containing }3\text{ vowels and }2\text{ consonants, can be}
\displaystyle \text{formed from the letters of the word INVOLUTE?}
\displaystyle \text{Answer:}
\displaystyle \text{The vowels in INVOLUTE are I, O, U and E.}
\displaystyle \text{The consonants are N, V, L and T.}
\displaystyle \text{Number of ways to select }3\text{ vowels from }4\text{ vowels}={}^{4}\mathrm{C}_{3}
\displaystyle \text{Number of ways to select }2\text{ consonants from }4\text{ consonants}={}^{4}\mathrm{C}_{2}
\displaystyle \text{The }5\text{ selected letters can be arranged in }5!\text{ ways.}
\displaystyle \therefore \text{Required number of words}
\displaystyle ={}^{4}\mathrm{C}_{3}\times{}^{4}\mathrm{C}_{2}\times5!
\displaystyle =4\times6\times120
\displaystyle =2880
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Find the number of permutations of }n\text{ different things taken }r\text{ at a time}
\displaystyle \text{such that }2\text{ specified things occur together.}
\displaystyle \text{Answer:}
\displaystyle \text{The }2\text{ specified things must be included in every selection.}
\displaystyle \therefore \text{The remaining }r-2\text{ things are selected from the other }n-2\text{ things.}
\displaystyle \text{Number of ways to select them}={}^{n-2}\mathrm{C}_{r-2}
\displaystyle \text{Treat the }2\text{ specified things as one unit.}
\displaystyle \text{Thus, the }r-2\text{ selected things and the one unit form }r-1\text{ objects.}
\displaystyle \text{These }r-1\text{ objects can be arranged in }(r-1)!\text{ ways.}
\displaystyle \text{The }2\text{ specified things can be arranged among themselves in }2!\text{ ways.}
\displaystyle \therefore \text{Required number of permutations}
\displaystyle ={}^{n-2}\mathrm{C}_{r-2}\times(r-1)!\times2!
\displaystyle =\frac{(n-2)!}{(r-2)!\,(n-r)!}\times(r-1)(r-2)!\times2
\displaystyle =\frac{2(r-1)(n-2)!}{(n-r)!}
\displaystyle =2(r-1)\,{}^{n-2}\mathrm{P}_{r-2}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Find the number of ways in which (a) a selection (b) an arrangement}
\displaystyle \text{of four letters can be made from the letters of the word PROPORTION.}
\displaystyle \text{Answer:}
\displaystyle \text{In the word PROPORTION, P occurs twice, R occurs twice and O occurs three times.}
\displaystyle \text{The letters T, I and N occur once each.}
\displaystyle \text{Thus, there are }6\text{ distinct kinds of letters: P, R, O, T, I and N.}
\displaystyle \text{(a) Number of selections}
\displaystyle \text{i) Three alike letters and one distinct letter}
\displaystyle \text{Only O occurs at least three times. The distinct letter can be chosen from the}
\displaystyle \text{other }5\text{ kinds of letters.}
\displaystyle \text{Number of selections}=1\times{}^{5}\mathrm{C}_{1}=5
\displaystyle \text{ii) Two alike letters of one kind and two alike letters of another kind}
\displaystyle \text{The repeated kinds can be chosen from P, R and O.}
\displaystyle \text{Number of selections}={}^{3}\mathrm{C}_{2}=3
\displaystyle \text{iii) Two alike letters and two distinct letters}
\displaystyle \text{The repeated kind can be chosen from P, R and O in }{}^{3}\mathrm{C}_{1}\text{ ways.}
\displaystyle \text{The two distinct kinds can then be chosen from the remaining }5\text{ kinds.}
\displaystyle \text{Number of selections}={}^{3}\mathrm{C}_{1}\times{}^{5}\mathrm{C}_{2}
\displaystyle =3\times10=30
\displaystyle \text{iv) All four letters are different}
\displaystyle \text{Number of selections}={}^{6}\mathrm{C}_{4}=15
\displaystyle \therefore \text{Total number of selections}=5+3+30+15
\displaystyle =53
\displaystyle \text{(b) Number of arrangements}
\displaystyle \text{i) Three alike letters and one distinct letter}
\displaystyle \text{Number of arrangements}=1\times{}^{5}\mathrm{C}_{1}\times\frac{4!}{3!\,1!}
\displaystyle =5\times4=20
\displaystyle \text{ii) Two alike letters of one kind and two alike letters of another kind}
\displaystyle \text{Number of arrangements}={}^{3}\mathrm{C}_{2}\times\frac{4!}{2!\,2!}
\displaystyle =3\times6=18
\displaystyle \text{iii) Two alike letters and two distinct letters}
\displaystyle \text{Number of arrangements}={}^{3}\mathrm{C}_{1}\times{}^{5}\mathrm{C}_{2}\times\frac{4!}{2!\,1!\,1!}
\displaystyle =3\times10\times12
\displaystyle =360
\displaystyle \text{iv) All four letters are different}
\displaystyle \text{Number of arrangements}={}^{6}\mathrm{C}_{4}\times4!
\displaystyle =15\times24
\displaystyle =360
\displaystyle \therefore \text{Total number of arrangements}=20+18+360+360
\displaystyle =758
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{How many words can be formed by taking }4\text{ letters at a time from the}
\displaystyle \text{letters of the word MORADABAD?}
\displaystyle \text{Answer:}
\displaystyle \text{In the word MORADABAD, A occurs }3\text{ times, D occurs }2\text{ times and}
\displaystyle \text{M, O, R and B occur once each.}
\displaystyle \text{i) Three alike letters and one distinct letter}
\displaystyle \text{Only A occurs three times. The remaining letter can be chosen from the other }5\text{ kinds.}
\displaystyle \text{Number of arrangements}=1\times{}^{5}\mathrm{C}_{1}\times\frac{4!}{3!\,1!}
\displaystyle =5\times4=20
\displaystyle \text{ii) Two alike letters of one kind and two alike letters of another kind}
\displaystyle \text{Only A and D can form two pairs.}
\displaystyle \text{Number of arrangements}={}^{2}\mathrm{C}_{2}\times\frac{4!}{2!\,2!}
\displaystyle =1\times6=6
\displaystyle \text{iii) Two alike letters and two distinct letters}
\displaystyle \text{The repeated letter can be A or D.}
\displaystyle \text{The remaining two distinct letters are chosen from the other }5\text{ kinds.}
\displaystyle \text{Number of arrangements}={}^{2}\mathrm{C}_{1}\times{}^{5}\mathrm{C}_{2}\times\frac{4!}{2!\,1!\,1!}
\displaystyle =2\times10\times12=240
\displaystyle \text{iv) All four letters are different}
\displaystyle \text{Number of arrangements}={}^{6}\mathrm{C}_{4}\times4!
\displaystyle =15\times24=360
\displaystyle \therefore \text{Total number of words}=20+6+240+360
\displaystyle =626
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{A businessman hosts a dinner for }21\text{ guests. He has }2\text{ round tables}
\displaystyle \text{which can accommodate }15\text{ and }6\text{ persons respectively. In how many ways}
\displaystyle \text{can he arrange the guests?}
\displaystyle \text{Answer:}
\displaystyle \text{Choose }15\text{ guests for the larger table.}
\displaystyle \text{Number of ways}={}^{21}\mathrm{C}_{15}
\displaystyle \text{These }15\text{ guests can be arranged around the round table in }(15-1)!\text{ ways.}
\displaystyle =14!\text{ ways.}
\displaystyle \text{The remaining }6\text{ guests can be arranged around the other round table in}
\displaystyle (6-1)!=5!\text{ ways.}
\displaystyle \therefore \text{Required number of arrangements}
\displaystyle ={}^{21}\mathrm{C}_{15}\times14!\times5!
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Find the number of combinations and permutations of }4\text{ letters taken}
\displaystyle \text{from the letters of the word EXAMINATION.}
\displaystyle \text{Answer:}
\displaystyle \text{In the word EXAMINATION, I and N occur twice each.}
\displaystyle \text{The letters E, X, A, M, T and O occur once each.}
\displaystyle \therefore \text{There are }8\text{ distinct kinds of letters.}
\displaystyle \text{The possible selections of }4\text{ letters are:}
\displaystyle \text{i) Two alike letters of one kind and two alike letters of another kind}
\displaystyle \text{Only I and N can form two pairs.}
\displaystyle \text{Number of combinations}={}^{2}\mathrm{C}_{2}=1
\displaystyle \text{Number of permutations}={}^{2}\mathrm{C}_{2}\times\frac{4!}{2!\,2!}
\displaystyle =1\times6=6
\displaystyle \text{ii) Two alike letters and two distinct letters}
\displaystyle \text{The repeated letter can be chosen from I and N in }{}^{2}\mathrm{C}_{1}\text{ ways.}
\displaystyle \text{The two distinct letters are chosen from the remaining }7\text{ kinds.}
\displaystyle \text{Number of combinations}={}^{2}\mathrm{C}_{1}\times{}^{7}\mathrm{C}_{2}
\displaystyle =2\times21=42
\displaystyle \text{Number of permutations}={}^{2}\mathrm{C}_{1}\times{}^{7}\mathrm{C}_{2}\times\frac{4!}{2!\,1!\,1!}
\displaystyle =2\times21\times12
\displaystyle =504
\displaystyle \text{iii) All four letters are different}
\displaystyle \text{Number of combinations}={}^{8}\mathrm{C}_{4}
\displaystyle =70
\displaystyle \text{Number of permutations}={}^{8}\mathrm{C}_{4}\times4!
\displaystyle =70\times24
\displaystyle =1680
\displaystyle \therefore \text{Total number of combinations}=1+42+70
\displaystyle =113
\displaystyle \therefore \text{Total number of permutations}=6+504+1680
\displaystyle =2190
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{A tea party is arranged for }16\text{ persons along two sides of a long table,}
\displaystyle \text{with }8\text{ chairs on each side. Four persons wish to sit on one particular side}
\displaystyle \text{and two on the other side. In how many ways can they be seated?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the two sides of the table be A and B.}
\displaystyle \text{The }4\text{ specified persons occupy }4\text{ seats on side A, and the }2\text{ specified persons}
\displaystyle \text{occupy }2\text{ seats on side B.}
\displaystyle \text{The remaining }10\text{ persons must fill the remaining }4\text{ seats on side A and}
\displaystyle \text{the remaining }6\text{ seats on side B.}
\displaystyle \text{Choose }4\text{ of the }10\text{ remaining persons for side A.}
\displaystyle \text{Number of ways}={}^{10}\mathrm{C}_{4}
\displaystyle \text{The remaining }6\text{ persons automatically occupy side B.}
\displaystyle \text{The }8\text{ persons on each side can be arranged in }8!\text{ ways.}
\displaystyle \therefore \text{Required number of arrangements}
\displaystyle ={}^{10}\mathrm{C}_{4}\times8!\times8!
\displaystyle ={}^{10}\mathrm{C}_{4}\times(8!)^2
\displaystyle \\

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