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Question: (i) If ${ }^{20} \mathrm{C}_{\mathrm{r}}={ }^{20} \mathrm{C}_{\mathrm{r}+6}$, find $\mathrm{r}$. (ii) If ${ }^{18} C_{r}={ }^{18} C_{r+2}$, find ${ }^{r} C_{5}$ Solution: Given: ${ }^{20} C_{r}={ }^{20} C_{r+6}$ To find: $r=?$ We know that: ${ }^{n} C_{r}={ }^{n} C_{n-r}$ $\Rightarrow{ }^{20} \mathrm{C}_{\mathrm{r}+6}={ }^{20} \mathrm{C}_{20-(\mathrm{r}+6)}$ $\Rightarrow{ }^{20} \mathrm{C}_{\mathrm{r}+6}={ }^{20} \mathrm{C}_{20-\mathrm{r}-6}={ }^{20} \mathrm{C}_{14-\mathrm{r}}$ $\Right...
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Question: (i) If ${ }^{n} C_{7}={ }^{n} C_{5}$, find $n$. (ii) If ${ }^{n} C_{14}={ }^{n} C_{16}$, find ${ }^{n} C_{28}$. (iii) If ${ }^{n} C_{16}={ }^{n} C_{14}$, find ${ }^{n} C_{27}$. Solution: (i) Given: ${ }^{n} C_{7}={ }^{n} C_{5}$ To find : $\mathrm{n}=?$ We know that: ${ }^{n} \mathrm{C}_{r}={ }^{n} \mathrm{C}_{n-r}$ $\Rightarrow{ }^{n} C_{7}={ }^{n} C_{n-7}$ $\Rightarrow{ }^{n} C_{n-7}={ }^{n} C_{5}$ $\Rightarrow n-7=5$ $\Rightarrow \mathrm{n}=7+5=12$ Ans: $\mathrm{n}=12$ (ii) Given: ${...
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Question: Verify that: (i) ${ }^{15} \mathrm{C}_{8}+{ }^{15} \mathrm{C}_{9}-{ }^{15} \mathrm{C}_{6}-{ }^{15} \mathrm{C}_{7}=0$ (ii) ${ }^{10} \mathrm{C}_{4}+{ }^{10} \mathrm{C}_{3}={ }^{11} \mathrm{C}_{4}$ Solution: (i) Given: ${ }^{15} \mathrm{C}_{8}+{ }^{15} \mathrm{C}_{9}-{ }^{15} \mathrm{C}_{6}-{ }^{15} \mathrm{C}_{7}$ To prove: ${ }^{15} \mathrm{C}_{8}+{ }^{15} \mathrm{C}_{9}-{ }^{15} \mathrm{C}_{6}-{ }^{15} \mathrm{C}_{7}=0$ We know that: ${ }^{n} C_{r}={ }^{n} C_{n-r}$ $\Rightarrow{ }^{15...
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Question: Evaluate: $\sum_{r=1}^{6}\left(\begin{array}{l}6 \\ r\end{array}\right)$ Solution: We know that: $\sum_{r=1}^{n}\left(\begin{array}{l}n \\ r\end{array}\right)=2^{n}-\left(\begin{array}{l}n \\ 0\end{array}\right)$ $\Rightarrow \sum_{r=1}^{6}\left(\begin{array}{l}6 \\ r\end{array}\right)=2^{6}-\left(\begin{array}{l}6 \\ 0\end{array}\right)$ $\Rightarrow \sum_{r=1}^{6}\left(\begin{array}{l}6 \\ r\end{array}\right)=64-1$ $\Rightarrow \sum_{r=1}^{6}\left(\begin{array}{l}6 \\ r\end{array}\ri...
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Question: Evaluate: $\sum_{r=1}^{6}\left(\begin{array}{l}6 \\ r\end{array}\right)$ Solution: We know that: $\sum_{r=1}^{n}\left(\begin{array}{l}n \\ r\end{array}\right)=2^{n}-\left(\begin{array}{l}n \\ 0\end{array}\right)$ $\Rightarrow \sum_{r=1}^{6}\left(\begin{array}{l}6 \\ r\end{array}\right)=2^{6}-\left(\begin{array}{l}6 \\ 0\end{array}\right)$ $\Rightarrow \sum_{r=1}^{6}\left(\begin{array}{l}6 \\ r\end{array}\right)=64-1$ $\Rightarrow \sum_{r=1}^{6}\left(\begin{array}{l}6 \\ r\end{array}\ri...
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Question: Evaluate: ${ }^{n+1} C_{n}$ Solution: We know that: ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{n !}{(n-r) ! \times r !}$ $\Rightarrow^{n+1} \mathrm{Cn}=\frac{(n+1) !}{(n+1-n) ! \times n !}$ $\Rightarrow^{n+1} C_{n}=\frac{(n+1) !}{1 ! \times n !}$ $\Rightarrow^{n+1} C_{n}=\frac{(n+1) !}{1 \times n !} \ldots(1 !=1)$ $\Rightarrow^{n+1} C_{n}=\frac{(n+1) \times n !}{1 \times n !}$ $\Rightarrow{ }^{n+1} C_{n}=\frac{(n+1)}{1}$ $\Rightarrow^{n+1} C_{n}=n+1$ Ans: ${ }^{n+1} C_{n}=n+1$...
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Question: ${ }^{71} \mathrm{C}_{71}$ Solution: We know that: ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{n !}{(n-r) ! \times r !}$ $\Rightarrow{ }^{71} \mathrm{C}_{71}=\frac{71 !}{(71-71) ! \times 71 !}$ $\Rightarrow{ }^{71} C_{71}=\frac{1}{0 !}$ $\Rightarrow{ }^{71} \mathrm{C}_{71}=\frac{1}{1} \ldots(0 !=1)$ $\Rightarrow{ }^{71} \mathrm{C}_{71}=1$ Ans: ${ }^{71} \mathrm{C}_{71}=1$...
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Question: Evaluate: ${ }^{90} \mathrm{C}_{88}$ Solution: We know that: ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{n !}{(n-r) ! \times r !}$ $\Rightarrow{ }^{90} \mathrm{C}_{88}=\frac{90 !}{(90-88) ! \times 88 !}$ $\Rightarrow{ }^{90} \mathrm{C}_{88}=\frac{90 !}{2 ! \times 88 !}$ $\Rightarrow{ }^{90} \mathrm{C}_{88}=\frac{90 \times 89 \times 88 !}{2 ! \times 88 !}$ $\Rightarrow{ }^{90} \mathrm{C}_{88}=\frac{90 \times 89}{2 !}$ $\Rightarrow{ }^{90} \mathrm{C}_{88}=\frac{90 \times 89}{2 \times...
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Question: Evaluate: ${ }^{16} \mathrm{C}_{13}$ Solution: We know that: ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{n !}{(n-r) ! \times r !}$ $\Rightarrow{ }^{16} \mathrm{C}_{13}=\frac{16 !}{(16-13) ! \times 13 !}$ $\Rightarrow{ }^{16} \mathrm{C}_{13}=\frac{16 !}{3 ! \times 13 !}$ $\Rightarrow{ }^{16} \mathrm{C}_{13}=\frac{16 \times 15 \times 14 \times 13 !}{3 ! \times 13 !}$ $\Rightarrow{ }^{16} \mathrm{C}_{13}=\frac{16 \times 15 \times 14}{3 !}$ $\Rightarrow{ }^{16} \mathrm{C}_{13}=\frac{16...
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Question: Evaluate: ${ }^{20} \mathrm{C}_{4}$ Solution: We know that: ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{n !}{(n-r) ! \times r !}$ $\Rightarrow{ }^{20} \mathrm{C}_{4}=\frac{20 !}{(20-4) ! \times 4 !}$ $\Rightarrow{ }^{20} \mathrm{C}_{4}=\frac{20 !}{16 ! \times 4 !}$ $\Rightarrow{ }^{20} \mathrm{C}_{4}=\frac{20 \times 19 \times 18 \times 17 \times 16 !}{16 ! \times 4 !}$ $\Rightarrow{ }^{20} \mathrm{C}_{4}=\frac{20 \times 19 \times 18 \times 17}{4 !}$ $\Rightarrow{ }^{20} \mathrm{C}_...
Read More →In how many ways can the letters of the word ‘PENCIL’
Question: In how many ways can the letters of the word PENCIL be arranged so that N is always next to E? Solution: Given: We have 6 letters To Find: Number of ways to arrange letters $P, E, N, C, I, L$ Condition: $N$ is always next to $E$ Here we need EN together in all arrangements. So, we will consider EN as a single letter. Now, we have 5 letters, i.e. $P, C, I, L$ and 'EN'. 5 letters can be arranged in ${ }^{5} P_{5}$ ways $\Rightarrow{ }^{5} P_{5}$ $\Rightarrow \frac{5 !}{(5-5) !}$ $\Righta...
Read More →A child has plastic toys bearing the digits 4, 4 and 5.
Question: A child has plastic toys bearing the digits 4, 4 and 5. How many 3-digit numbers can he make using them? Solution: Given: We have toys with bearing 4, 4 and 5 To Find: Number of 3-digit numbers he can make The formula used: The number of permutations of $n$ objects, where $p_{1}$ objects are of one kind, $p_{2}$ are of the second kind, ..., $p_{k}$ is of a $k^{\text {th }}$ kind and the rest, if any, are of a different kind is $=\frac{\mathrm{n} !}{\mathrm{p}_{1} ! \mathrm{p}_{2} ! \ld...
Read More →In how many ways can 5 boys and 3 girls be seated in a row so
Question: In how many ways can 5 boys and 3 girls be seated in a row so that each girl is between 2 boys? Solution: Given: We have 5 boys and 3 girls To Find: Number of ways of seating so that 5 boys and 3 girls are seated in a row and each girl is between 2 boys The formula used: The number of permutations of n different objects taken r at a time (object does not repeat) is ${ }^{n} P_{r}=\frac{n !}{(n-r) !}$ The only arrangement possible is Number of ways for boys $={ }^{n} P_{r}$ $={ }^{5} P_...
Read More →How many 5-digit numbers can be formed by using the digits 0, 1 and 2?
Question: How many 5-digit numbers can be formed by using the digits 0, 1 and 2? Solution: Given: We have 3 digits, i.e. 0, 1 and 2 To Find: Number of 5-digit numbers formed Let us represent the arrangement For forming a 5-digit number, we have to fill 5 vacant spaces. But the first place cannot be filled with 0 , hence for filling first place, we have only 1 and 2 First place can be filled with 2 numbers, i.e. $1,2=2$ ways Second place can be filled with 3 numbers $=3$ ways Third place can be f...
Read More →In how many ways can 4 women draw water from 4 taps if no tap remains unused?
Question: In how many ways can 4 women draw water from 4 taps if no tap remains unused? Solution: Given: We have 4 women and 4 taps To Find: Number of ways of drawing water Condition: No tap remains unused Let us represent the arrangement The first woman can use any of the four taps $=4$ Ways The second woman can use the remaining three taps $=3$ ways The third woman can use the remaining two taps $=2$ ways The fourth woman can use the remaining one tap $=1$ way Total number of ways $=4 \times 3...
Read More →In how many ways can 4 letters be posted in 5 letter boxes?
Question: In how many ways can 4 letters be posted in 5 letter boxes? Solution: Given: We have 4 letters and 5 letter boxes To Find: Number of ways of posting letters. One letter can be posted in any of 5 letter boxes. We have to assume that all the letters are different. So for first letter i.e. $L_{1}$, we have 5 ways Similarly for, $L_{2}=5$ ways $L_{3}=5$ ways $L_{4}=5$ ways Total number of ways $=5 \times 5 \times 5 \times 5=625$ In 625 ways 4 letters can be posted in 5 letter boxes....
Read More →How many words can be formed by the letters of the word ‘SUNDAY’?
Question: How many words can be formed by the letters of the word SUNDAY? Solution: Given: We have 6 letters To Find: Number of words formed with Letter of the word 'SUNDAY:' 'SUNDAY' consist of 6 letters. NOTE: - Unless specified, assume that repetition is not allowed. Let us represent the arrangement with an example 6 ways 5 ways 4 ways 3 ways 2 ways 1 way We have 6 places First place can be filled with 6 letters, i.e. S,U,N,D,A,Y = 6 way Second place can be filled with 5 letters (as one lette...
Read More →How many permutations of the letters of the word ‘APPLE’ are there?
Question: How many permutations of the letters of the word APPLE are there? Solution: Given: We have 5 letters To Find: Number of words formed with Letter of the word APPLE. The formula used: The number of permutations of $n$ objects, where $p_{1}$ objects are of one kind, $p_{2}$ are of the second kind, ..., $p_{k}$ is of a $k^{\text {th }}$ kind and the rest if any, are of a different kind is $=\frac{n !}{p_{1} ! p_{2} ! \ldots \ldots \ldots \ldots p_{k} !}$ 'APPLE' consists of 5 letters out o...
Read More →How many different words can be formed by using all the letters of the word
Question: How many different words can be formed by using all the letters of the word ALLAHABAD? Solution: Given: We have 9 letters To Find: Number of words formed with Letter of the word 'ALLAHABAD.' The formula used: The number of permutations of $n$ objects, where $p_{1}$ objects are of one kind, $p_{2}$ are of the second kind, ..., $p_{k}$ is of a $k^{\text {th }}$ kind and the rest if any, are of a different kind is $=\frac{n !}{p_{1} ! p_{2} ! \ldots \ldots \ldots \ldots p_{k} !}$ ALLAHABA...
Read More →In how many ways can the letters of the word ‘PERMUTATIONS’
Question: In how many ways can the letters of the word PERMUTATIONS be arranged if each word starts with P and ends with S? Solution: Given: We have 12 letters To Find: Number of words formed with Letter of the word PERMUTATIONS. The formula used: The number of permutations of $n$ objects, where $p_{1}$ objects are of one kind, $\mathrm{p}_{2}$ are of the second kind, ..., $\mathrm{p}_{\mathrm{k}}$ is of $\mathrm{a} \mathrm{k}^{\text {th }}$ kind and the rest if any, are of a different kind is $...
Read More →In how many ways can the letters of the word ‘CHEESE’ be arranged?
Question: In how many ways can the letters of the word CHEESE be arranged? Solution: Given: We have 6 letters To Find: Number of words formed with Letter of the word CHEESE. The formula used: The number of permutations of $n$ objects, where $p_{1}$ objects are of one kind, $p_{2}$ are of the second kind, ..., $p_{k}$ is of a $k^{\text {th }}$ kind and the rest if any, are of a different kind is $=\frac{n !}{p_{1} ! p_{2} ! \ldots \ldots \ldots \ldots \ldots p_{k} !}$ Suppose we have these words ...
Read More →How many numbers divisible by 5 and lying between 4000 and 5000 can be
Question: How many numbers divisible by 5 and lying between 4000 and 5000 can be formed from the digits 4, 5, 6, 7, 8 if repetition of digits is allowed? Solution: Given: We have 5 digits, i.e. 4,5,6,7,8 To Find: Number of numbers divisible by 5 Condition: (i) Number should be between 4000 and 5000 (ii) Repetition of digits is allowed Here as the number is lying between 4000 and 5000 , we can conclude that the number is of 4 -digits and the number must be starting with 4 . Now, for a number to b...
Read More →How many 3-digit numbers above 600 can be formed by using the digits
Question: How many 3-digit numbers above 600 can be formed by using the digits 2, 3, 4, 5, 6, if repetition of digits is allowed? Solution: Given: We have 5 digits i.e. 2,3,4,5,6 To Find: Number of 3-digit numbers Condition: (i) Number should be greater than 600 (ii) Repetition of digits is allowed For forming a 3 digit number, we have to fill 3 vacant spaces. But as the number should be above 600 , hence the first place must be occupied with 6 only because no other number is greater than 6 . Le...
Read More →How many 3-digit numbers are there with no digit repeated?
Question: How many 3-digit numbers are there with no digit repeated? Solution: Given: We have 10 numbers i.e. 0,1,2,3,4,5,6,7,8,9 To Find: Number of 3-digit numbers formed with no repetition of digits. Conditions: No digit is repeated Let us represent the 3-digit number First place can be filled with 9 numbers, i.e. $1,2,3,4,5,6,7,8,9$ ( 0 cannot be placed as it will make it a 2-digit number) $=9$ ways Second place can be filled with remaining 9 numbers (as one number is used already) = 9 ways S...
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Question: if$\frac{1}{4 !}+\frac{1}{5 !}=\frac{x}{6 !}$, find the value of $x$ Solution: To Find: Value of n Given: $\frac{1}{4 !}+\frac{1}{5 !}=\frac{x}{6 !}$ Formula Used: $n !=(n) \times(n-1) \times(n-2) \times(n-3)$ $3 \times 2 \times 1$ Now, $\frac{1}{4 !}+\frac{1}{5 !}=\frac{x}{6 !}$ $\Rightarrow \frac{1}{24}+\frac{1}{120}=\frac{x}{720}(4 !=24,5 !=120)$ $\Rightarrow \frac{5+1}{120}=\frac{x}{720}$ $\Rightarrow \frac{6}{120}=\frac{x}{720}$ $\Rightarrow x=36$...
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