In how many ways can 6 pictures be hung from 4 picture nails on a wall?
Question: In how many ways can 6 pictures be hung from 4 picture nails on a wall? Solution: To find: number of ways of hanging 6 pictures on 4 picture nails. There are 6 pictures to be placed in 4 places. Formula: Number of permutations of $n$ distinct objects among $r$ different places, where repetition is not allowed, is $P(n, r)=n ! /(n-r) !$ Therefore, a permutation of 6 different objects in 4 places is $P(6,4)=\frac{\frac{6 !}{(6-4) !}}{2 !}=\frac{6 !}{2 !}=\frac{720}{2}=360$ This can be do...
Read More →If there are 6 periods on each working day of a school, in how many ways
Question: If there are 6 periods on each working day of a school, in how many ways can one arrange 5 subjects such that each subject is allowed at least one period? Solution: To find: number of ways of arranging 5 subjects in 6 periods. Condition: at least 1 period for each subject. 5 subjects in 6 periods can be arranged in $P(6,5)$. Remaining 1 period can be arranged in $\mathrm{P}(5,1)$ Formula: Number of permutations of $n$ distinct objects among $r$ different places, where repetition is not...
Read More →Ten students are participating in a race. In how many ways can the first
Question: Ten students are participating in a race. In how many ways can the first three prizes be won? Solution: To find: number of ways of winning the first three prizes The first price can go to any of the 10 students. The second price can go to any of the remaining 9 students. The third price can go to any of the remaining 8 students. Formula: Number of permutations of $n$ distinct objects among $r$ different places, where repetition is not allowed, is $P(n, r)=n ! /(n-r) !$ Therefore, a per...
Read More →Five letters F, K, R, R and V one in each were purchased from a plastic
Question: Five letters F, K, R, R and V one in each were purchased from a plastic warehouse. How many ordered pairs of letters, to be used as initials, can be formed from them? Solution: (i) The number of initials is 1 In this case, all letters have one chance (i.e. letters $F, K, R, V$ ). Formula: Number of permutations of $n$ distinct objects among $r$ different places, where repetition is not allowed, is $P(n, r)=n ! /(n-r) !$ Therefore, a permutation of 4 different objects in 1 place is $P(4...
Read More →There are 6 items in column A and 6 items in column B.
Question: There are 6 items in column A and 6 items in column B. A student is asked to match each item in column A with an item in column B. How many possible, correct or incorrect answers are there to this question? Solution: To find: number of possibilities of a selection of answers Each item in column $\mathrm{A}$ can select another item in column $\mathrm{B}$. Therefore the question involves selecting each item from column $\mathrm{A}$ to each item in column $B$. this can be done in $P(6,6)$...
Read More →It is required to seat 5 men and 3 women in a row so that the women occupy
Question: It is required to seat 5 men and 3 women in a row so that the women occupy the even places. How many such arrangements are possible? Solution: To find: number of arrangements in which women sit in even places Condition: women occupy even places Here the total number of people is 8 . In this question first, the arrangement of women is required. The positions where women can be made to sit is $2^{\text {nd }}, 4^{\text {th }}, 6^{\text {th }}, 8^{\text {th }}$. There are 4 even places in...
Read More →Six students are contesting the election for the president ship of the
Question: Six students are contesting the election for the president ship of the students, union. In how many ways can their names be listed on the ballot papers? Solution: To find: number of arrangements of names on a ballot paper. There are six contestants contesting in the elections. Name of any 1 student out of six can appear first on the ballot paper. 2 position on the ballot paper can be filled by rest of the five names and so on. Formula: Number of permutations of $n$ distinct objects amo...
Read More →In how many ways can 4 different books, one each in chemistry, physics,
Question: In how many ways can 4 different books, one each in chemistry, physics, biology and mathematics, be arranged on a shelf? Solution: To find: number of arrangements of 4 different books in a shelf. There are 4 different books. Any one of the four different books can be placed on the shelf first. Similarly, in the next position, 1 book out of 3 can be placed. Finally, the last book will have a single place to fit. Formula: Number of permutations of $n$ distinct objects among $r$ different...
Read More →In how many ways can 6 women draw water from 6 wells if no well remains
Question: In how many ways can 6 women draw water from 6 wells if no well remains unused? Solution: To find: number of arrangements of 6 women drawing water from 6 wells Here, 6 wells are needed to be used by 6 women. Therefore any one of the 6 women can draw water from the 1 well. Similarly, any 5 women can draw water from the $2^{\text {nd }}$ well and so on. Lastly, there will be single women left to draw water from the $6^{\text {th }}$ well. Formula: Number of permutations of $n$ distinct o...
Read More →In how many ways can 5 children stand in a queue?
Question: In how many ways can 5 children stand in a queue? Solution: To find: number of arrangements of 5 children in a queue. Here, 5 places are needed to be occupied by 5 children. Therefore any one of the 5 children can occupy first place. Similarly, any 4 children can occupy second place and so on. Lastly, there will be a single person to occupy the 5 position Formula: Number of permutations of $n$ distinct objects among $r$ different places, where repetition is not allowed, is $P(n, r)=n !...
Read More →In how many ways can 7 people line up at a ticket window of a cinema hall?
Question: In how many ways can 7 people line up at a ticket window of a cinema hall? Solution: To find: number of arrangements of 7 people in a queue. Here there are 7 spaces to be occupied by 7 people. Therefore 7 people can occupy first place. Similarly, 6 people can occupy second place and so on. Lastly, there will be a single person to occupy the 7 positions. Formula: Number of permutations of $n$ distinct objects among $r$ different places, where repetition is not allowed, is $P(n, r)=n ! /...
Read More →In how many ways can 5 persons occupy 3 vacant seats?
Question: In how many ways can 5 persons occupy 3 vacant seats? Solution: To find: number of arrangements of 5 people in 3 seats. Consider three seats $\underline{A} \underline{B} \underline{C}$ Now, place A can be occupied by any 1 person out of $5 .$ Then place B can be occupied by any 1 person from remaining 4 and for $C$ there are 3 people to occupy the seat. Formula: Number of permutations of n distinct objects among r different places, where repetition is not allowed, is $P(n, r)=n ! /(n-r...
Read More →Find the number of permutations of 10 objects, taken 4 at a time.
Question: Find the number of permutations of 10 objects, taken 4 at a time. Solution: To find: the number of permutations of 10 objects, taken 4 at a time. Formula Used: Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by, ${ }_{n P_{r}}=\frac{n !}{(n-r) !}$ ${ }_{10 p_{4}}=\frac{10 !}{6 !}$ ${ }^{10} \mathrm{P}_{4}={ }_{10} \times 9 \times 8 \times 7$ ${ }^{10} \mathrm{P}_{4}=5040$ Hence, the number of permutations of 10 objects, tak...
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Question: Prove that $1+1 .{ }^{1} P_{1}+2 .{ }^{2} P_{2}+3 .{ }^{3} P_{3}+\ldots . n .{ }^{n} P_{n}={ }^{n+1} P_{n+1}$ Solution: To Prove: $1+1 .{ }^{1} \mathrm{P}_{1}+2 .{ }^{2} \mathrm{P}_{2}+3 .{ }^{3} \mathrm{P}_{3}+\ldots . . \mathrm{n} .{ }^{n} \mathrm{P}_{\mathrm{n}}={ }^{\mathrm{n}+1} \mathrm{P}_{\mathrm{n}+1}$ Formula Used: Total number of ways in which $\mathrm{n}$ objects can be arranged in $\mathrm{r}$ places (Such that no object is replaced) is given by, ${ }_{n}{ }_{P_{r}}=\frac{n...
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Question: Find $n$, If ${ }^{n+5} P_{n+1}=\frac{11}{2}(n-1) \cdot{ }^{n+3} P_{n}$, find $n .$ Solution: To find: the value of n Formula Used: Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by, ${ }_{n} P_{r}=\frac{n !}{(n-r) !}$ ${ }^{n+5} P_{n+1}=\frac{11}{2}(n-1) \cdot{ }^{n+3} P_{n}$ $\frac{(n+5) !}{4 !}=\frac{11}{2}(n-1) \frac{(n+3) !}{3 !}$ $\frac{(n+5)(n+4)(n+3) !}{4 \times 3 !}=\frac{11}{2}(n-1) \frac{(n+3) !}{3 !}$ $\frac{(n...
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Question: If ${ }^{2 n-1} P_{n}:{ }^{2 n+1} P_{n-1},=22: 7$, find $n .$ Solution: To find: the value of n Formula Used: Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by, ${ }_{n p_{r}}=\frac{n !}{(n-r) !}$ ${ }^{2 n-1} P_{n}:{ }^{2 n+1} P_{n-1},=22: 7$ $\frac{(2 n-1) !}{(n-1) !}: \frac{(2 n+1) !}{(n+2) !}=\frac{22}{7}$ $\frac{(2 n-1) !}{(n-1) !}: \frac{(2 n+1)(2 n)(2 n-1) !}{(n+2)(n+1) n(n-1) !}=\frac{22}{7}$ $\frac{(2 n-1) !}{(n-1...
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Question: If ${ }^{15} \operatorname{Pr}-1:{ }^{16} \operatorname{Pr}-2,=3: 4$, find $\mathrm{r}$. Solution: To find: the value of r Formula Used: Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by, ${ }_{n p_{r}}=\frac{n !}{(n-r) !}$ ${ }^{15} P_{r-1}:{ }^{16} P_{r-2},=3: 4$ $\frac{15 !}{(16-r) !}: \frac{16 !}{(18-r) !}=\frac{3}{4}$ $\frac{15 !}{(16-\mathrm{r}) !}: \frac{16 \times 15 !}{(18-\mathrm{r})(17-\mathrm{r})(16-\mathrm{r}) ...
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Question: (i) If ${ }^{n} P_{4}:{ }^{n} P_{5},=1: 2$, find $n$. (ii) If ${ }^{n-1} P_{3}:{ }^{n+1} P_{3},=5: 12$, find $n$. Solution: To find: the value of n Formula Used: Total number of ways in which $\mathrm{n}$ objects can be arranged in $\mathrm{r}$ places (Such that no object is replaced) is given by, ${ }_{{ }^{\mathrm{n}} \operatorname{Pr}}=\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) !}$ ${ }^{\mathrm{n}} \mathrm{P}_{4}:{ }^{\mathrm{n}} \mathrm{P}_{5}=1: 2$ $\frac{n !}{(n-4) !}: \frac{n ...
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Question: (i) If If ${ }^{5} P_{r}=2 \times{ }^{6} P_{r-1}$, find $r$. (ii) If ${ }^{20} P_{r}=13 \times{ }^{20} P_{r-1}$, find $r$. (iii) If ${ }^{11} P_{r}={ }^{12} P_{r-1}$, find $r$. Solution: (i) To find: the value of $r$ Formula Used: Total number of ways in which $n$ objects can be arranged in r places (Such that no object is replaced) is given by, ${ }_{n} P_{r}=\frac{n !}{(n-r) !}$ ${ }^{5} P_{r}=2^{\times}{ }^{6} P_{r-1}$ $\frac{5 !}{(5-\mathrm{r}) !}=\left(2 \times \frac{6 !}{(7-\math...
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Question: (i) If ${ }^{n} P_{5}=20 \times{ }^{n} P_{3}$, find $n$. (ii) If $16 \times{ }^{n} P_{3}=13 \times{ }^{n+1} P_{3}$, find $n$. (iii) If ${ }^{2 n} P_{3}=100 \times{ }^{n} P_{2}$, find $n$. Solution: (i) To find: the value of $n$ Formula Used: Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by, ${ }_{n P_{r}}=\frac{n !}{(n-r) !}$ ${ }^{n} P_{5}=20 \times{ }^{n} P_{3}$ $\frac{n !}{(n-5) !}=\left(20 \times \frac{n !}{(n-3) !}\r...
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Question: Prove that ${ }^{9} p_{3}+3 \times{ }^{9} p_{2}={ }^{10} p_{3}$ Solution: To Prove: ${ }^{9} \mathrm{P}_{3}+3 \times{ }^{9} \mathrm{P}_{2}={ }^{10} \mathrm{P}_{3}$ Formula Used: Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by, ${ }_{n} P_{r}=\frac{n !}{(n-r) !}$ The equation given below needs to be proved i.e ${ }^{9} P_{3}+3 \times{ }^{9} P_{2}={ }^{10} P_{3}$ $\frac{9 !}{(9-3) !}_{+}\left(3 \times \frac{9 !}{(9-2) !}\r...
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Question: Evaluate: ${ }^{9} \mathrm{P}_{0}$ Solution: To find: the value of ${ }^{9} \mathrm{P}_{0}$ Formula Used: Total number of ways in which $\mathrm{n}$ objects can be arranged in $\mathrm{r}$ places (Such that no object is replaced) is given by, ${ }_{n} P_{r}=\frac{n !}{(n-r) !}$ Therefore, ${ }_{9} \mathrm{P}_{0}=\frac{9 !}{(9-0) !}$ ${ }^{9} \mathrm{P}_{0}=1$ Thus, the value of ${ }^{9} \mathrm{P}_{0}$ is 1 ....
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Question: Evaluate: ${ }^{6} \mathrm{P}_{6}$ Solution: To find: the value of ${ }^{6} \mathrm{P}_{6}$ Formula Used: Total number of ways in which $n$ objects can be arranged in $r$ places (Such that no object is replaced) is given by, ${ }_{n P_{r}}=\frac{n !}{(n-r) !}$ Therefore, ${ }_{6} \mathrm{P}_{6}=\frac{6 !}{(6-6) !}$ ${ }^{62} P_{3}=6 \times 5 \times 4 \times 3 \times 2 \times 1=720$ Thus, the value of ${ }^{6} \mathrm{P}_{6}$ is 720 ....
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Question: Evaluate: ${ }^{62} P_{3}$ Solution: To find: the value of ${ }^{62} \mathrm{P}_{3}$ Formula Used: Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by, ${ }_{n} P_{r}=\frac{n !}{(n-r) !}$ Therefore, ${ }_{62} \mathrm{P}_{3}=\frac{62 !}{(62-3) !}$ ${ }^{62} \mathrm{P}_{3}=62 \times 61 \times 60 \times 59=226920$ Thus, the value of ${ }^{62} \mathrm{P}_{3}$ is 226920 ....
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Question: Evaluate: ${ }^{10} \mathrm{P}_{4}$ Solution: To find: the value of ${ }^{10} \mathrm{P}_{4}$ Formula Used: Total number of ways in which $n$ objects can be arranged in $r$ places (Such that no object is replaced) is given by, ${ }_{n p_{r}}=\frac{n !}{(n-r) !}$ Therefore, ${ }_{10 p_{4}}=\frac{10 !}{(10-4) !}$ ${ }^{10} \mathrm{P}_{4}=10 \times 9 \times 8 \times 7$ ${ }^{10} \mathrm{P}_{4}=5040$ Thus, the value of ${ }^{10} \mathrm{P}_{4}$ is 5040 ....
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