\displaystyle \textbf{Question 1: } \text{Evaluate each of the following:}
\displaystyle \text{i) }{}^8P_3\qquad \text{ii) }{}^{10}P_4\qquad \text{iii) }{}^6P_6\qquad \text{iv) }P(6,4)
\displaystyle \text{Answer:}
\displaystyle \text{We know, }{}^nP_r=\frac{n!}{(n-r)!}.
\displaystyle \text{i) }{}^8P_3=\frac{8!}{(8-3)!}=\frac{8!}{5!}=\frac{8\times7\times6\times5!}{5!}=8\times7\times6=336.
\displaystyle \text{ii) }{}^{10}P_4=\frac{10!}{(10-4)!}=\frac{10!}{6!}=\frac{10\times9\times8\times7\times6!}{6!}=10\times9\times8\times7=5040.
\displaystyle \text{iii) }{}^6P_6=\frac{6!}{(6-6)!}=\frac{6!}{0!}=6!=720.
\displaystyle \text{iv) }P(6,4)={}^6P_4=\frac{6!}{(6-4)!}=\frac{6!}{2!}=\frac{6\times5\times4\times3\times2!}{2!}=6\times5\times4\times3=360.
\displaystyle \\

\displaystyle \textbf{Question 2: } \text{If }P(5,r)=P(6,r-1),\text{ find }r.
\displaystyle \text{Answer:}
\displaystyle P(5,r)=P(6,r-1)
\displaystyle \Rightarrow \frac{5!}{(5-r)!}=\frac{6!}{[6-(r-1)]!}
\displaystyle \Rightarrow \frac{5!}{(5-r)!}=\frac{6\times5!}{(7-r)!}
\displaystyle \Rightarrow \frac{1}{(5-r)!}=\frac{6}{(7-r)(6-r)(5-r)!}
\displaystyle \Rightarrow (7-r)(6-r)=6
\displaystyle \Rightarrow 42-13r+r^2=6
\displaystyle \Rightarrow r^2-13r+36=0
\displaystyle \Rightarrow (r-9)(r-4)=0
\displaystyle \Rightarrow r=9\text{ or }r=4
\displaystyle \text{Since }P(5,r)\text{ requires }0\leq r\leq5,\text{ the value }r=9\text{ is inadmissible.}
\displaystyle \therefore r=4.
\displaystyle \\

\displaystyle \textbf{Question 3: } \text{If }5P(4,n)=6\cdot P(5,n-1),\text{ find }n.
\displaystyle \text{Answer:}
\displaystyle 5P(4,n)=6\cdot P(5,n-1)
\displaystyle \Rightarrow 5\cdot{}^4P_n=6\cdot{}^5P_{n-1}
\displaystyle \Rightarrow 5\cdot\frac{4!}{(4-n)!}=6\cdot\frac{5!}{[5-(n-1)]!}
\displaystyle \Rightarrow \frac{120}{(4-n)!}=\frac{720}{(6-n)!}
\displaystyle \Rightarrow \frac{1}{(4-n)!}=\frac{6}{(6-n)!}
\displaystyle \Rightarrow \frac{1}{(4-n)!}=\frac{6}{(6-n)(5-n)(4-n)!}
\displaystyle \Rightarrow (6-n)(5-n)=6
\displaystyle \Rightarrow 30-11n+n^2=6
\displaystyle \Rightarrow n^2-11n+24=0
\displaystyle \Rightarrow (n-3)(n-8)=0
\displaystyle \Rightarrow n=3\text{ or }n=8
\displaystyle \text{Since }P(4,n)\text{ requires }0\leq n\leq4,\text{ the value }n=8\text{ is inadmissible.}
\displaystyle \therefore n=3.
\displaystyle \\

\displaystyle \textbf{Question 4: } \text{If }P(n,5)=20\cdot P(n,3),\text{ find }n.
\displaystyle \text{Answer:}
\displaystyle P(n,5)=20\cdot P(n,3)
\displaystyle \Rightarrow {}^nP_5=20\cdot{}^nP_3
\displaystyle \Rightarrow \frac{n!}{(n-5)!}=20\cdot\frac{n!}{(n-3)!}
\displaystyle \Rightarrow \frac{1}{(n-5)!}=\frac{20}{(n-3)(n-4)(n-5)!}
\displaystyle \Rightarrow (n-3)(n-4)=20
\displaystyle \Rightarrow n^2-7n+12=20
\displaystyle \Rightarrow n^2-7n-8=0
\displaystyle \Rightarrow (n-8)(n+1)=0
\displaystyle \Rightarrow n=8\text{ or }n=-1
\displaystyle \text{Since }P(n,5)\text{ requires }n\geq5,\text{ the value }n=-1\text{ is inadmissible.}
\displaystyle \therefore n=8.
\displaystyle \\

\displaystyle \textbf{Question 5: } \text{If }{}^nP_4=360,\text{ find the value of }n.
\displaystyle \text{Answer:}
\displaystyle {}^nP_4=360
\displaystyle \Rightarrow \frac{n!}{(n-4)!}=360
\displaystyle \Rightarrow n(n-1)(n-2)(n-3)=360
\displaystyle \Rightarrow n(n-1)(n-2)(n-3)=6\times5\times4\times3
\displaystyle \text{For }n=6,\quad {}^6P_4=6\times5\times4\times3=360.
\displaystyle \text{Also, }{}^nP_4\text{ increases as }n\text{ increases for }n\geq4.
\displaystyle \therefore n=6.
\displaystyle \\

\displaystyle \textbf{Question 6: } \text{If }P(9,r)=360,\text{ find }r.
\displaystyle \text{Answer:}
\displaystyle P(9,r)=360
\displaystyle \Rightarrow {}^9P_r=360
\displaystyle \Rightarrow \frac{9!}{(9-r)!}=360
\displaystyle \Rightarrow \frac{9!}{(9-r)!}=9\times8\times7\times6
\displaystyle \Rightarrow \frac{9!}{(9-r)!}=\frac{9!}{5!}
\displaystyle \Rightarrow 9-r=5
\displaystyle \therefore r=4.
\displaystyle \\

\displaystyle \textbf{Question 7: } \text{If }P(11,r)=P(12,r-1),\text{ find }r.
\displaystyle \text{Answer:}
\displaystyle P(11,r)=P(12,r-1)
\displaystyle \Rightarrow \frac{11!}{(11-r)!}=\frac{12!}{[12-(r-1)]!}
\displaystyle \Rightarrow \frac{11!}{(11-r)!}=\frac{12\times11!}{(13-r)!}
\displaystyle \Rightarrow \frac{1}{(11-r)!}=\frac{12}{(13-r)(12-r)(11-r)!}
\displaystyle \Rightarrow (13-r)(12-r)=12
\displaystyle \Rightarrow 156-25r+r^2=12
\displaystyle \Rightarrow r^2-25r+144=0
\displaystyle \Rightarrow (r-9)(r-16)=0
\displaystyle \Rightarrow r=9\text{ or }r=16
\displaystyle \text{Since }P(11,r)\text{ requires }0\le r\le11,\text{ the value }r=16\text{ is inadmissible.}
\displaystyle \therefore r=9.
\displaystyle \\

\displaystyle \textbf{Question 8: } \text{If }P(n,4)=12\cdot P(n,2),\text{ find }n.
\displaystyle \text{Answer:}
\displaystyle P(n,4)=12\cdot P(n,2)
\displaystyle \Rightarrow \frac{n!}{(n-4)!}=12\cdot\frac{n!}{(n-2)!}
\displaystyle \Rightarrow \frac{1}{(n-4)!}=\frac{12}{(n-2)(n-3)(n-4)!}
\displaystyle \Rightarrow (n-2)(n-3)=12
\displaystyle \Rightarrow n^2-5n+6=12
\displaystyle \Rightarrow n^2-5n-6=0
\displaystyle \Rightarrow (n-6)(n+1)=0
\displaystyle \Rightarrow n=6\text{ or }n=-1
\displaystyle \text{Since }P(n,4)\text{ requires }n\geq4,\text{ the value }n=-1\text{ is inadmissible.}
\displaystyle \therefore n=6.
\displaystyle \\

\displaystyle \textbf{Question 9: } \text{If }P(n-1,3):P(n,4)=1:9,\text{ find }n.
\displaystyle \text{Answer:}
\displaystyle P(n-1,3):P(n,4)=1:9
\displaystyle \Rightarrow \frac{P(n-1,3)}{P(n,4)}=\frac{1}{9}
\displaystyle \Rightarrow \frac{\frac{(n-1)!}{[(n-1)-3]!}}{\frac{n!}{(n-4)!}}=\frac{1}{9}
\displaystyle \Rightarrow \frac{(n-1)!}{(n-4)!}\times\frac{(n-4)!}{n!}=\frac{1}{9}
\displaystyle \Rightarrow \frac{(n-1)!}{n!}=\frac{1}{9}
\displaystyle \Rightarrow \frac{1}{n}=\frac{1}{9}
\displaystyle \therefore n=9.
\displaystyle \\

\displaystyle \textbf{Question 10: } \text{If }P(2n-1,n):P(2n+1,n-1)=22:7,\text{ find }n.
\displaystyle \text{Answer:}
\displaystyle P(2n-1,n):P(2n+1,n-1)=22:7
\displaystyle \Rightarrow \frac{(2n-1)!}{[(2n-1)-n]!}:\frac{(2n+1)!}{[(2n+1)-(n-1)]!}=22:7
\displaystyle \Rightarrow \frac{(2n-1)!}{(n-1)!}:\frac{(2n+1)!}{(n+2)!}=22:7
\displaystyle \Rightarrow \frac{(2n-1)!}{(n-1)!}\times\frac{(n+2)!}{(2n+1)!}=\frac{22}{7}
\displaystyle \Rightarrow \frac{(2n-1)!}{(n-1)!}\times\frac{(n+2)(n+1)n(n-1)!}{(2n+1)(2n)(2n-1)!}=\frac{22}{7}
\displaystyle \Rightarrow \frac{n(n+1)(n+2)}{2n(2n+1)}=\frac{22}{7}
\displaystyle \Rightarrow \frac{(n+1)(n+2)}{2(2n+1)}=\frac{22}{7}
\displaystyle \Rightarrow 7(n+1)(n+2)=44(2n+1)
\displaystyle \Rightarrow 7(n^2+3n+2)=88n+44
\displaystyle \Rightarrow 7n^2+21n+14=88n+44
\displaystyle \Rightarrow 7n^2-67n-30=0
\displaystyle \Rightarrow (n-10)(7n+3)=0
\displaystyle \Rightarrow n=10\text{ or }n=-\frac{3}{7}
\displaystyle \text{Since }n\text{ must be a non-negative integer, }n=-\frac{3}{7}\text{ is inadmissible.}
\displaystyle \therefore n=10.
\displaystyle \\

\displaystyle \textbf{Question 11: } \text{If }P(n,5):P(n,3)=2:1,\text{ find }n.
\displaystyle \text{Answer:}
\displaystyle P(n,5):P(n,3)=2:1
\displaystyle \Rightarrow \frac{n!}{(n-5)!}:\frac{n!}{(n-3)!}=2:1
\displaystyle \Rightarrow \frac{n!}{(n-5)!}\times\frac{(n-3)!}{n!}=\frac{2}{1}
\displaystyle \Rightarrow \frac{(n-3)(n-4)(n-5)!}{(n-5)!}=2
\displaystyle \Rightarrow (n-3)(n-4)=2
\displaystyle \Rightarrow n^2-7n+12=2
\displaystyle \Rightarrow n^2-7n+10=0
\displaystyle \Rightarrow (n-5)(n-2)=0
\displaystyle \Rightarrow n=5\text{ or }n=2
\displaystyle \text{Since }P(n,5)\text{ requires }n\geq5,\text{ the value }n=2\text{ is inadmissible.}
\displaystyle \therefore n=5.
\displaystyle \\

\displaystyle \textbf{Question 12: } \text{Prove that:}
\displaystyle 1\cdot P(1,1)+2\cdot P(2,2)+3\cdot P(3,3)+\ldots+n\cdot P(n,n)
\displaystyle =P(n+1,n+1)-1.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=1\cdot P(1,1)+2\cdot P(2,2)+3\cdot P(3,3)+\ldots+n\cdot P(n,n)
\displaystyle =1\cdot{}^1P_1+2\cdot{}^2P_2+3\cdot{}^3P_3+\ldots+n\cdot{}^nP_n
\displaystyle =1\cdot\frac{1!}{(1-1)!}+2\cdot\frac{2!}{(2-2)!}+3\cdot\frac{3!}{(3-3)!}+\ldots+n\cdot\frac{n!}{(n-n)!}
\displaystyle =1\cdot1!+2\cdot2!+3\cdot3!+\ldots+n\cdot n!
\displaystyle =\sum\limits_{r=1}^{n}r\cdot r!
\displaystyle =\sum\limits_{r=1}^{n}[(r+1)-1]r!
\displaystyle =\sum\limits_{r=1}^{n}[(r+1)!-r!]
\displaystyle =(2!-1!)+(3!-2!)+(4!-3!)+\ldots+[(n+1)!-n!]
\displaystyle =(n+1)!-1!
\displaystyle =(n+1)!-1
\displaystyle =P(n+1,n+1)-1=\text{RHS}.
\displaystyle \therefore \text{The given identity is proved.}
\displaystyle \\

\displaystyle \textbf{Question 13: } \text{If }P(15,r-1):P(16,r-2)=3:4,\text{ find }r.
\displaystyle \text{Answer:}
\displaystyle P(15,r-1):P(16,r-2)=3:4
\displaystyle \Rightarrow {}^{15}P_{r-1}:{}^{16}P_{r-2}=3:4
\displaystyle \Rightarrow \frac{15!}{[15-(r-1)]!}:\frac{16!}{[16-(r-2)]!}=3:4
\displaystyle \Rightarrow \frac{15!}{(16-r)!}\times\frac{(18-r)!}{16!}=\frac{3}{4}
\displaystyle \Rightarrow \frac{15!}{(16-r)!}\times\frac{(18-r)(17-r)(16-r)!}{16\cdot15!}=\frac{3}{4}
\displaystyle \Rightarrow \frac{(18-r)(17-r)}{16}=\frac{3}{4}
\displaystyle \Rightarrow (18-r)(17-r)=12
\displaystyle \Rightarrow r^2-35r+306=12
\displaystyle \Rightarrow r^2-35r+294=0
\displaystyle \Rightarrow (r-14)(r-21)=0
\displaystyle \Rightarrow r=14\text{ or }r=21
\displaystyle \text{Since }P(15,r-1)\text{ requires }r-1\leq15,\text{ the value }r=21\text{ is inadmissible.}
\displaystyle \therefore r=14.
\displaystyle \\

\displaystyle \textbf{Question 14: } \text{If }{}^{n+5}P_{n+1}=\left[\frac{11(n-1)}{2}\right]{}^{n+3}P_n,\text{ find }n.
\displaystyle \text{Answer:}
\displaystyle {}^{n+5}P_{n+1}=\left[\frac{11(n-1)}{2}\right]{}^{n+3}P_n
\displaystyle \Rightarrow \frac{(n+5)!}{[(n+5)-(n+1)]!}=\left[\frac{11(n-1)}{2}\right]\frac{(n+3)!}{[(n+3)-n]!}
\displaystyle \Rightarrow \frac{(n+5)!}{4!}=\left[\frac{11(n-1)}{2}\right]\frac{(n+3)!}{3!}
\displaystyle \Rightarrow \frac{(n+5)(n+4)(n+3)!}{24}=\frac{11(n-1)(n+3)!}{12}
\displaystyle \Rightarrow (n+5)(n+4)=22(n-1)
\displaystyle \Rightarrow n^2+9n+20=22n-22
\displaystyle \Rightarrow n^2-13n+42=0
\displaystyle \Rightarrow (n-6)(n-7)=0
\displaystyle \therefore n=6\text{ or }n=7.
\displaystyle \\

\displaystyle \textbf{Question 15: } \text{In how many ways can five children stand in a queue?}
\displaystyle \text{Answer:}
\displaystyle \text{We need to arrange }5\text{ children out of }5.
\displaystyle \therefore \text{Required number of ways}={}^5P_5=\frac{5!}{(5-5)!}=5!=120.
\displaystyle \\

\displaystyle \textbf{Question 16: } \text{From among the }36\text{ teachers in a school, one principal and one}
\displaystyle \text{vice-principal are to be appointed. In how many ways can this be done?}
\displaystyle \text{Answer:}
\displaystyle \text{We need to permute }2\text{ teachers out of }36.
\displaystyle \therefore \text{Required number of ways}={}^{36}P_2=\frac{36!}{(36-2)!}=36\times35=1260.
\displaystyle \\

\displaystyle \textbf{Question 17: } \text{Four letters }E,\ K,\ S\text{ and }V\text{ are available, one of each.}
\displaystyle \text{How many ordered pairs of letters, to be used as initials, can be formed?}
\displaystyle \text{Answer:}
\displaystyle \text{We need to permute }2\text{ letters out of }4.
\displaystyle \therefore \text{Required number of ways}={}^4P_2=\frac{4!}{(4-2)!}=4\times3=12.
\displaystyle \\

\displaystyle \textbf{Question 18: } \text{Four books, one each in Chemistry, Physics, Biology and Mathematics,}
\displaystyle \text{are to be arranged on a shelf. In how many ways can this be done?}
\displaystyle \text{Answer:}
\displaystyle \text{We need to arrange }4\text{ books out of }4.
\displaystyle \therefore \text{Required number of ways}={}^4P_4=\frac{4!}{(4-4)!}=4!=24.
\displaystyle \\

\displaystyle \textbf{Question 19: } \text{Find the number of different }4\text{-letter words, with or without}
\displaystyle \text{meanings, that can be formed from the letters of the word }\text{`NUMBER'.}
\displaystyle \text{Answer:}
\displaystyle \text{The word }\text{NUMBER}\text{ has }6\text{ distinct letters.}
\displaystyle \text{We need to permute }4\text{ letters out of }6.
\displaystyle \therefore \text{Required number of ways}={}^6P_4=\frac{6!}{(6-4)!}=6\times5\times4\times3=360.
\displaystyle \\

\displaystyle \textbf{Question 20: } \text{How many three-digit numbers are there with distinct digits,}
\displaystyle \text{if each digit is odd?}
\displaystyle \text{Answer:}
\displaystyle \text{The odd digits are }1,\,3,\,5,\,7,\,9.
\displaystyle \text{We need to permute }3\text{ digits out of }5.
\displaystyle \therefore \text{Required number of ways}={}^5P_3=\frac{5!}{(5-3)!}=5\times4\times3=60.
\displaystyle \\

\displaystyle \textbf{Question 21: } \text{How many words, with or without meaning, can be formed by using all}
\displaystyle \text{the letters of the word }\text{`DELHI', using each letter exactly once?}
\displaystyle \text{Answer:}
\displaystyle \text{The word }\text{DELHI}\text{ has }5\text{ distinct letters.}
\displaystyle \text{We need to permute }5\text{ letters out of }5.
\displaystyle \therefore \text{Required number of ways}={}^5P_5=\frac{5!}{(5-5)!}=5!=120.
\displaystyle \\

\displaystyle \textbf{Question 22: } \text{How many words, with or without meaning, can be formed by using all}
\displaystyle \text{the letters of the word }\text{`TRIANGLE'?}
\displaystyle \text{Answer:}
\displaystyle \text{The word }\text{TRIANGLE}\text{ has }8\text{ distinct letters.}
\displaystyle \text{We need to permute }8\text{ letters out of }8.
\displaystyle \therefore \text{Required number of ways}={}^8P_8=\frac{8!}{(8-8)!}=8!=40320.
\displaystyle \\

\displaystyle \textbf{Question 23: } \text{There are two works each of }3\text{ volumes and two works each of }2
\displaystyle \text{volumes. In how many ways can the }10\text{ books be placed on a shelf so that}
\displaystyle \text{the volumes of the same work are not separated?}
\displaystyle \text{Answer:}
\displaystyle \text{Treat each work as a single block. There are }4\text{ such blocks.}
\displaystyle \text{The }4\text{ blocks can be arranged in }{}^4P_4=4!\text{ ways.}
\displaystyle \text{Within the blocks, the volumes can be arranged in }(3!\times3!)\times(2!\times2!).
\displaystyle \therefore \text{Required number of ways}=4!\times(3!\times3!)\times(2!\times2!)=3456.
\displaystyle \\

\displaystyle \textbf{Question 24: } \text{There are }6\text{ items in Column }A\text{ and }6\text{ items in Column }B.
\displaystyle \text{A student is asked to match each item in Column }A\text{ with an item in}
\displaystyle \text{Column }B.\text{ How many possible, correct or incorrect, answers are there?}
\displaystyle \text{Answer:}
\displaystyle \text{Each item in Column }A\text{ is to be matched with a distinct item in Column }B.
\displaystyle \text{Hence the }6\text{ items of Column }B\text{ can be arranged in }6!\text{ ways.}
\displaystyle \therefore \text{Required number of matchings}={}^6P_6=6!=720.
\displaystyle \\

\displaystyle \textbf{Question 25: } \text{How many three-digit numbers are there with no digit repeated?}
\displaystyle \text{Answer:}
\displaystyle \text{Total arrangements of }3\text{ distinct digits chosen from }10\text{ digits}={}{}^{10}P_3
\displaystyle =\frac{10!}{(10-3)!}=\frac{10!}{7!}=10\times9\times8=720.
\displaystyle \text{Among these, those beginning with }0\text{ are }{}^9P_2
\displaystyle =\frac{9!}{(9-2)!}=\frac{9!}{7!}=9\times8=72.
\displaystyle \therefore \text{Required number of three-digit numbers}=720-72=648.
\displaystyle \\

\displaystyle \textbf{Question 26: } \text{How many }6\text{-digit telephone numbers can be constructed using the}
\displaystyle \text{digits }0,1,2,3,4,5,6,7,8,9\text{ if each number starts with }35\text{ and no digit}
\displaystyle \text{appears more than once?}
\displaystyle \text{Answer:}
\displaystyle \text{The first two digits are fixed as }3\text{ and }5.
\displaystyle \text{Therefore, the remaining }4\text{ positions are to be filled using }4\text{ of the remaining }8\text{ digits.}
\displaystyle \therefore \text{Required number of telephone numbers}={}^8P_4
\displaystyle =\frac{8!}{(8-4)!}=\frac{8!}{4!}=8\times7\times6\times5=1680.
\displaystyle \\

\displaystyle \textbf{Question 27: } \text{In how many ways can }6\text{ boys and }5\text{ girls be arranged for a group}
\displaystyle \text{photograph if the girls are to sit on chairs in a row and the boys are to stand}
\displaystyle \text{in a row behind them?}
\displaystyle \text{Answer:}
\displaystyle \text{The }5\text{ girls can be arranged in a row in }5!\text{ ways.}
\displaystyle \text{The }6\text{ boys can be arranged in a row in }6!\text{ ways.}
\displaystyle \therefore \text{Required number of arrangements}=5!\times6!
\displaystyle =120\times720=86400.
\displaystyle \\

\displaystyle \textbf{Question 28: } \text{If }a\text{ denotes the number of permutations of }(x+2)\text{ things}
\displaystyle \text{taken all at a time, }b\text{ the number of permutations of }x\text{ things taken }11
\displaystyle \text{at a time and }c\text{ the number of permutations of }(x-11)\text{ things taken all}
\displaystyle \text{at a time such that }a=182bc,\text{ find the value of }x.
\displaystyle \text{Answer:}
\displaystyle a={}^{x+2}P_{x+2}=\frac{(x+2)!}{[(x+2)-(x+2)]!}=\frac{(x+2)!}{0!}=(x+2)!
\displaystyle b={}^xP_{11}=\frac{x!}{(x-11)!}
\displaystyle c={}^{x-11}P_{x-11}=\frac{(x-11)!}{[(x-11)-(x-11)]!}=\frac{(x-11)!}{0!}=(x-11)!
\displaystyle \text{Given, }a=182bc
\displaystyle \Rightarrow (x+2)!=182\left[\frac{x!}{(x-11)!}\right](x-11)!
\displaystyle \Rightarrow (x+2)!=182x!
\displaystyle \Rightarrow (x+2)(x+1)x!=182x!
\displaystyle \Rightarrow (x+2)(x+1)=182
\displaystyle \Rightarrow x^2+3x+2=182
\displaystyle \Rightarrow x^2+3x-180=0
\displaystyle \Rightarrow (x-12)(x+15)=0
\displaystyle \Rightarrow x=12\text{ or }x=-15
\displaystyle \text{Since }P(x,11)\text{ requires }x\geq11,\text{ the value }x=-15\text{ is inadmissible.}
\displaystyle \therefore x=12.
\displaystyle \\

\displaystyle \textbf{Question 29: } \text{How many }3\text{-digit numbers can be formed using the digits }1\text{ to }9
\displaystyle \text{if no digit is repeated?}
\displaystyle \text{Answer:}
\displaystyle \text{We need to arrange }3\text{ digits out of the }9\text{ available digits.}
\displaystyle \therefore \text{Required number of }3\text{-digit numbers}={}^9P_3
\displaystyle =\frac{9!}{(9-3)!}=\frac{9!}{6!}=9\times8\times7=504.
\displaystyle \\

\displaystyle \textbf{Question 30: } \text{How many }3\text{-digit even numbers can be formed using the digits}
\displaystyle 1,2,3,4,5,6,7\text{ if no digit is repeated?}
\displaystyle \text{Answer:}
\displaystyle \text{The units digit must be even. The possible choices are }2,4\text{ and }6.
\displaystyle \text{If the units digit is }2,\text{ the remaining two places can be filled in }{}^6P_2=6\times5=30\text{ ways.}
\displaystyle \text{Similarly, if the units digit is }4,\text{ the number of arrangements is }30.
\displaystyle \text{Similarly, if the units digit is }6,\text{ the number of arrangements is }30.
\displaystyle \therefore \text{Required number of }3\text{-digit even numbers}=30+30+30=90.
\displaystyle \\

\displaystyle \textbf{Question 31: } \text{Find the number of }4\text{-digit numbers that can be formed using the}
\displaystyle \text{digits }1,2,3,4,5\text{ if no digit is repeated. How many of these will be even?}
\displaystyle \text{Answer:}
\displaystyle \text{The number of }4\text{-digit numbers that can be formed}={}^5P_4
\displaystyle =\frac{5!}{(5-4)!}=\frac{5!}{1!}=5\times4\times3\times2=120.
\displaystyle \text{For an even number, the units digit must be }2\text{ or }4.
\displaystyle \text{If the units digit is }2,\text{ the remaining three places can be filled in }{}^4P_3=24\text{ ways.}
\displaystyle \text{If the units digit is }4,\text{ the remaining three places can be filled in }{}^4P_3=24\text{ ways.}
\displaystyle \therefore \text{Number of even numbers}=24+24=48.
\displaystyle \\

\displaystyle \textbf{Question 32: } \text{All the letters of the word }\text{`EAMCOT'}\text{ are arranged in different}
\displaystyle \text{possible ways. Find the number of arrangements in which no two vowels are}
\displaystyle \text{adjacent to each other.}
\displaystyle \text{Answer:}
\displaystyle \text{The consonants are }M,C,T\text{ and the vowels are }E,A,O.
\displaystyle \text{The }3\text{ consonants can be arranged in }3!=6\text{ ways.}
\displaystyle \text{For each arrangement of consonants, there are }4\text{ gaps:}
\displaystyle \_\,C\,\_\,C\,\_\,C\,\_
\displaystyle \text{To ensure that no two vowels are adjacent, the }3\text{ vowels must occupy }3\text{ of these }4\text{ gaps.}
\displaystyle \text{The vowels can be placed in the }4\text{ gaps in }{}^4P_3=\frac{4!}{(4-3)!}=24\text{ ways.}
\displaystyle \therefore \text{Required number of arrangements}=3!\times{}^4P_3=6\times24=144.
\displaystyle \\

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