Find the number of different signals that can be generated by arranging at
Question: Find the number of different signals that can be generated by arranging at least 2 flags in order (one below the other) on a vertical staff, if five different flags are available. Solution: Given:5 Flags Way of generating signal using 2 different flags $={ }^{5} \mathrm{P}_{2}$ (way of selecting 2 things out of 5 things with considering arrangement.) Way of generating signal using 3 different flags $={ }^{5} \mathrm{P}_{3}$ Way of generating signal using 4 different flags $={ }^{5} \ma...
Read More →ac and the outcomes are recorded.
Question: (ac and the outcomes are recorded. How many possible outcomes are there? (b) How many possible outcomes if the coin is tossed. (i) four times? (ii) five times? (iii) $\mathrm{n}$ times? Solution: (a) A coin is tossed three times So possible number of outcomes $=2^{3}=8$ (HHH,HHT,HTH,HTT,THH,THT,TTH,TTT) (b) i) A coin is tossed four times So possible number of outcomes $=2^{4}=16$ (HHHH,HHHT,HHTH,HHTT,HTHH,HTHT,HTTH,HTTT, THHH,THHT,THTH,THTT,TTHH ,TTHT,TTTH,TTTT) (ii) A coin is tossed n...
Read More →How many 8-digit telephone numbers can be constructed using the digits 0
Question: How many 8-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 270 and n digit appears more than once? Solution: Given:8 digit telephone number starts with 270 . To find: How many 8-digit telephone numbers can be constructed? There are 10 digits between 0 to 9,and three of them are utilized in filling up the first three digits i.e.270 of the 8 digit phone number, so remaining number of digits=10- $3=7$, and this need to be used in filling up th...
Read More →In how many ways can a vowel, a consonant and a digit be chosen out of the
Question: In how many ways can a vowel, a consonant and a digit be chosen out of the 26 letters of the English alphabet and the 10 digits? Solution: To find: number of ways in which a vowel, a consonant and a digit be chosen out of the 26 letters of the English alphabet and the 10 digits. e.g. Way of selecting a vowel from 5 vowels $={ }^{5} C_{1}$ Way of selecting a consonant from 26 consonants $={ }^{21} \mathrm{C}_{1}$ Way of selecting a digit from 10 digits $={ }^{10} \mathrm{C}_{1}$ So ways...
Read More →In a school, there are four sections of 40 students each in XI standard.
Question: In a school, there are four sections of 40 students each in XI standard. In how many ways can a set of 4 student representatives be chosen, one from each section? Solution: Given: there are four sections of 40 students each in XI standard. To find : number of ways in which a set of 4 student representatives be chosen, one from each section. Ways of selecting one student from section $1:{ }^{40} \mathrm{C}_{1}$ Ways of selecting one student from section $2:{ }^{40} \mathrm{C}_{1}$ Ways ...
Read More →In a textbook on mathematics there are three exercises A, B and C
Question: In a textbook on mathematics there are three exercises A, B and C consisting of 12, 18 and 10 questions respectively. In how many ways can three questions be selected choosing one from each exercise? Solution: Given: three exercises A, B and C consisting of 12, 18 and 10 questions respectively To find: number of ways in which three questions be selected choosing one from each exercise. Ways of selecting one question from exercise $A:{ }^{12} C_{1}$ (way of selecting one element from n ...
Read More →In How many ways can 5 ladies draw water from 5 taps, assuming the no tap
Question: In How many ways can 5 ladies draw water from 5 taps, assuming the no tap remains unused? Solution: To find: number of ways in which 5 ladies draw water from 5 taps. Condition: no tap remains unused The condition given is that no well should remain unused. So possible number of ways are: $5 \times 4 \times 3 \times 2 \times 1=120$....
Read More →In How many ways can 4 people be seated in a row containing 5 seats?
Question: In How many ways can 4 people be seated in a row containing 5 seats? Solution: To find : Number of ways in which 4 people can be seated in a row containing 5 seats. The possible number of ways in which 4 people be seated in a row containing 5 seats $={ }^{7} \mathrm{P}_{4}$ (There are 5 places to be filled with 4 persons where arrangement doesn't matter.) ${ }^{7} \mathrm{P}_{4}=\frac{7 !}{(7-4) !} \ldots\left({ }^{\mathrm{n}} \mathrm{P}_{\mathrm{r}}=\frac{n !}{(n-r) !}\right)$ $=\frac...
Read More →There are 12 steamers plying between A and B.
Question: There are 12 steamers plying between A and B. In how many ways could the round trip from A be made if the return was made on \ (i) the same steamer? (ii) a different steamer? Solution: Given: 12 steamers plying between A and B. To find: number of ways the round trip from A can be made. (i) The steamer which will go from A to B will be returning back, since the given condition is that same steamer should return. There are 12 steamers available so there are 12 different ways to make arou...
Read More →A, B and C are three cities. There are 5 routes from A to B and 3 routes from
Question: A, B and C are three cities. There are 5 routes from A to B and 3 routes from B to C. How many different routes are there from A to C via B? Solution: Given: 5 routes from A to B and 3 routes from B to C. To find: number of different routes from $A$ to $C$ via $B$. Let $E_{1}$ be the event : 5 routes from A to B Let $E_{2}$ be the event : 3 routes from $B$ to $C$ Since going from $A$ to $C$ via $B$ is only possible if both the events $E_{1}$ and $E_{2}$ occur simultaneously. So there a...
Read More →There are 10 buses running between Delhi and Agra.
Question: There are 10 buses running between Delhi and Agra. In how many ways can a man go from Delhi to Agra and return by a different bus? Solution: Given: 10 buses running between Delhi and Agra. To Find: Number of ways a man can go from Delhi to Agra and return by a different bus. There are 10 buses running between Delhi and Agra so there are 10 different ways to go from Delhi to Agra. The man cannot return from the same bus he went so number of ways are reduced to 9. These second event occu...
Read More →Prove that
Question: Prove that (i) $\frac{n !}{r !}=n(n-1)(n-2) \ldots(r+1)$ (ii) $(n-r+1)$ $\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}+1) !}=\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) !}$ (iii) $\frac{\mathrm{n} !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !}+\frac{\mathrm{n} !}{(\mathrm{r}-1) !(\mathrm{n}-\mathrm{r}+1) !}=\frac{(\mathrm{n}+1) !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}+1) !}$ Solution: (i) To Prove $\frac{n !}{r !}=n(n-1)(n-2) \ldots(r+1)$ Formula: $n !=n \times(n-1) !$ L.H.S. $=\frac{n !}{r !}$ W...
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Question: Prove that (i) $\frac{n !}{r !}=n(n-1)(n-2) \ldots(r+1)$ (ii) $(n-r+1)$ $\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}+1) !}=\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) !}$ (iii) $\frac{\mathrm{n} !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !}+\frac{\mathrm{n} !}{(\mathrm{r}-1) !(\mathrm{n}-\mathrm{r}+1) !}=\frac{(\mathrm{n}+1) !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}+1) !}$ Solution: (i) To Prove $\frac{n !}{r !}=n(n-1)(n-2) \ldots(r+1)$ Formula: $n !=n \times(n-1) !$ L.H.S. $=\frac{n !}{r !}$ W...
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Question: Prove that $(n+2) \times(n !)+(n+1) !=(n !) \cdot(2 n+3)$ Solution: To Prove: $(n+2) \times(n !)+(n+1) !=(n !) \times(2 n+3)$ Formula: $n !=n \times(n-1) !$ L.H.S. $=(n+2) \times(n !)+(n+1) !$ $=(n+2) \times(n !)+(n+1) \times(n !)$ $=(n !) \times[(n+2)+(n+1)]$ $=(n !) \times(2 n+3)$ $=$ R.H.S. $\therefore$ L.H.S. $=$ R.H.S. Conclusion : $(n+2) \times(n !)+(n+1) !=(n !) \times(2 n+3)$...
Read More →Evaluate
Question: Evaluate $\frac{\mathrm{n} !}{(\mathrm{r} !) \times(\mathrm{n}-\mathrm{r}) !}$ when $n=15$ and $r=12$ Solution: Given : n = 15 and r = 12 To Find: Value of $\frac{n !}{(r !) \times(n-r) !}$ at given $\mathrm{n}$ and $\mathrm{r}$ Formula : $n !=n \times(n-1) !$ $n !=n \times(n-1) \times(n-2) \ldots \ldots \ldots \ldots 3 \times 2 \times 1$ Let , $x=\frac{n !}{(r !) \times(n-r) !}$ Substituting n = 15 and r = 12 in above equation, $\therefore x=\frac{(15 !)}{(12 !) \times(15-12) !}$ $\th...
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Question: If $\frac{(2 n) !}{(3 !) \times(2 n-3) !}: \frac{n !}{(2 !) \times(n-2) !}=44: 3$, find the value of $n$ Solution: Given Equation : $\frac{(2 n) !}{(3 !) \times(2 n-3) !}: \frac{n !}{(2 !) \times(n-2) !}=44: 3$ To Find : Value of n Formula: $n !=n \times(n-1) !$ By given equation $\frac{(2 n) !}{(3 !) \times(2 n-3) !}: \frac{n !}{(2 !) \times(n-2) !}=44: 3$ $\therefore \frac{(2 n) !}{\frac{(3 !) \times(2 n-3) !}{n !}}=\frac{44}{3}$ $\therefore \frac{(2 n) !}{(3 !) \times(2 n-3) !} \tim...
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Question: If $\frac{\mathrm{n} !}{(2 !) \times(\mathrm{n}-2) !}: \frac{\mathrm{n} !}{(4 !) \times(\mathrm{n}-4) !}=2: 1$, find the value of $\mathrm{n}$ Solution: Given Equation : $\frac{n !}{(2 !) \times(n-2) !}: \frac{n !}{(4 !) \times(n-4) !}=2: 1$ To Find : Value of n Formula $n !=n \times(n-1) !$ By given equation, $\frac{n !}{(2 !) \times(n-2) !}: \frac{n !}{(4 !) \times(n-4) !}=2: 1$ $\therefore \frac{\frac{n !}{(2 !) \times(n-2) !}}{\frac{n !}{(4 !) \times(n-4) !}}=\frac{2}{1}$ $\therefo...
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Question: If $(n+3) !=56 \times(n+1) !$, find the value of $n .$ Solution: Given Equation : $(n+3) !=56 \times(n+1) !$ To Find : Value of $n$ Formula: $n !=n \times(n-1) !$ By given equation, $(n+3) !=56 \times(n+1) !$ By using above formula we can write, $\therefore(n+3) \times(n+2) \times(n+1) !=56 \times(n+1) !$ Cancelling the term $(n+1) !$ from both the sides, $\therefore(n+3) \times(n+2)=56$ $\therefore(n+3) \times(n+2)=(8) \times(7)$ Comparing both the sides, we get $\therefore \mathrm{n}...
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Question: If $(n+2) !=2550 \times n !$, find the value of $n$ Solution: Given Equation : $(n+2) !=2550 \times n !$ To Find : Value of n Formula: $n !=n \times(n-1) !$ By given equation, $(n+2) !=2550 \times n !$ By using above formula we can write, $\therefore(n+2) \times(n+1) \times(n !)=2550 \times n !$ Cancelling the term (n)! from both the sides $\therefore(n+2) \times(n+1)=2550$ $\therefore(n+2) \times(n+1)=(51) \times(50)$ Comparing both the sides, we get, $\therefore \mathrm{n}=49$ Conclu...
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Question: If $(n+1) !=12 \times(n-1) !$, find the value of $n$. Solution: Given Equation : $(n+1) !=12 \times(n-1) !$ To Find : Value of n Formula: $n !=n \times(n-1) !$ By given equation, $(n+1) !=12 \times(n-1) !$ By using above formula we can write, $\therefore(n+1) \times(n) \times(n-1) !=12 \times(n-1) !$ Cancelling the term $(n-1) !$ from both the sides, $\therefore(n+1) \times(n)=12 \ldots \ldots \ldots$ eq(1) $\therefore(n+1) \times(n)=(4) \times(3)$ Comparing both the sides, we get, $\t...
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Question: If $(n+1) !=12 \times(n-1) !$, find the value of $n$. Solution: Given Equation : $(n+1) !=12 \times(n-1) !$ To Find : Value of n Formula: $n !=n \times(n-1) !$ By given equation, $(n+1) !=12 \times(n-1) !$ By using above formula we can write, $\therefore(n+1) \times(n) \times(n-1) !=12 \times(n-1) !$ Cancelling the term $(n-1) !$ from both the sides, $\therefore(n+1) \times(n)=12 \ldots \ldots \ldots$ eq(1) $\therefore(n+1) \times(n)=(4) \times(3)$ Comparing both the sides, we get, $\t...
Read More →Which of the following are true of false?
Question: Which of the following are true of false? (i) $(2+3) !=2 !+3 !$ (ii) $(2 \times 3) !=(2 !) \times(3 !)$ Solution: Option (i) and (ii) both are false Proofs : For option (i) L.H.S. $=(2+3) !=(5 !)=120$ R.H.S. $=(2 !)+(3 !)=2+6=8$ $\therefore$ L.H.S. \neqR.H.S. For option (ii), L.H.S. $=(2 \times 3) !=(6 !)=720$ R.H.S $=(2 !) \times(3 !)=4 \times 6=24$ $\therefore$ L.H.S. $\neq$ R.H.S. Important Notes : for any two whole numbers a and b, - $(a+b) ! \neq(a !)+(b !)$ $\cdot(a \times b) ! \...
Read More →Write the following products in factorial notation:
Question: Write the following products in factorial notation: (i) $6 \times 7 \times 8 \times 9 \times 10 \times 11 \times 12$ (ii) $3 \times 6 \times 9 \times 12 \times 15$ Solution: (i) Formula : $n !=n \times(n-1) \times(n-2) \ldots \ldots \ldots \ldots 3 \times 2 \times 1$ Let $x=12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6$ Multiplying and dividing by $(5 \times 4 \times 3 \times 2 \times 1)$ $\therefore x=\frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times...
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Question: If $\frac{1}{6 !}+\frac{1}{7 !}=\frac{x}{8 !^{\prime}}$, find the value of $x$ Solution: Given Equation : $\frac{1}{6 !}+\frac{1}{7 !}=\frac{x}{8 !}$ To Find : Value of x Formula: $n !=n \times(n-1) !$ By given equation $\frac{1}{6 !}+\frac{1}{7 !}=\frac{x}{8 !}$ $\therefore \frac{8 \times 7}{8 \times 7 \times 6 !}+\frac{8}{8 \times 7 !}=\frac{x}{8 !}$ By using the above formula we can write, $\therefore \frac{56}{8 !}+\frac{8}{8 !}=\frac{x}{8 !}$ $\therefore \frac{64}{8 !}=\frac{x}{8 ...
Read More →Prove that
Question: Prove that $\frac{1}{10 !}+\frac{1}{11 !}+\frac{1}{12 !}=\frac{145 !}{12 !}$ Solution: To Prove : $\frac{1}{10 !}+\frac{1}{11 !}+\frac{1}{12 !}=\frac{145}{12 !}$ Formula : $n !=n \times(n-1) !$ $L . H . S .=\frac{1}{10 !}+\frac{1}{11 !}+\frac{1}{12 !}$ $=\frac{12 \times 11}{12 \times 11 \times(10 !)}+\frac{12}{12 \times(11 !)}+\frac{1}{12 !}$ $=\frac{132}{12 !}+\frac{12}{12 !}+\frac{1}{12 !}$ $=\frac{145}{12 !}$ $=\mathrm{R} \cdot \mathrm{H} \cdot \mathrm{S} .$ $\therefore \mathrm{L} \...
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