Evaluate :
Question: Evaluate : $(\sqrt{2}+1)^{6}+(\sqrt{2}-1)^{6}$ Solution: To find: Value of $(\sqrt{2}+1)^{6}+(\sqrt{2}-1)^{6}$ Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$ (ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$ $(a+1)^{6}=$ $\left[{ }^{6} \mathrm{C}_{0} \mathrm{a}^{6}\right]+\left[{ }^{6} \mathrm{C}_{1} \mathrm{a}^{6-1} 1\right]+\left[{ }^{6} \mathrm{C}_{2} \mathrm{a}^{6-2} 1^{2}...
Read More →Using binomial theorem, expand each of the following:
Question: Using binomial theorem, expand each of the following: $\left(3 x^{2}-2 a x+3 a^{2}\right)^{3}$ Solution: To find: Expansion of $\left(3 x^{2}-2 a x+3 a^{2}\right)^{3}$ Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$ (ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$ We have, $\left(3 x^{2}-2 a x+3 a^{2}\right)^{3}$ Let, $\left(3 x^{2}-2 a x\right)=p \ldots$ (i) The equation beco...
Read More →Using binomial theorem, expand each of the following:
Question: Using binomial theorem, expand each of the following: $\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}, x \neq 0$ Solution: To find: Expansion of $\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}, x \neq 0$ Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$ (ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$ We have, $\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}, x \neq 0$ Let $\left(1+\frac{x}{...
Read More →Using binomial theorem, expand each of the following:
Question: Using binomial theorem, expand each of the following: $\left(1+2 x-3 x^{2}\right)^{4}$ Solution: To find: Expansion of $\left(1+2 x-3 x^{2}\right)^{4}$ Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$ (ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$ We have, $\left(1+2 x-3 x^{2}\right)^{4}$ Let $(1+2 x)=a$ and $\left(-3 x^{2}\right)=b \ldots$ (i) Now the equation becomes $(a+b)...
Read More →Using binomial theorem, expand each of the following:
Question: Using binomial theorem, expand each of the following: $(\sqrt[3]{x}-\sqrt[3]{y})^{6}$ Solution: To find: Expansion of $(\sqrt[3]{x}-\sqrt[3]{y})^{6}$ Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$ (ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$ We have, $(\sqrt[3]{x}-\sqrt[3]{y})^{6}$ We can write $\sqrt[3]{x}$ as $x^{\frac{1}{3}}$ and $\sqrt[3]{y}$ as $y^{\frac{1}{3}}$ Now,...
Read More →Using binomial theorem, expand each of the following:
Question: Using binomial theorem, expand each of the following: $(\sqrt{x}+\sqrt{y})^{8}$ Solution: To find: Expansion of $(\sqrt{x}+\sqrt{y})^{8}$ Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$ (ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$ We have, $(\sqrt{x}+\sqrt{y})^{8}$ We can write $\sqrt{x}$ as $x^{\frac{1}{2}}$ and $\sqrt{y}$ as $y^{\frac{1}{2}}$ Now, we have to solve for $\...
Read More →Using binomial theorem, expand each of the following:
Question: Using binomial theorem, expand each of the following: $\left(x-\frac{1}{y}\right)^{5}$ Solution: To find: Expansion of $\left(x-\frac{1}{y}\right)^{5}$ Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$ (ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$ We have, $\left(x-\frac{1}{y}\right)^{5}$ $\Rightarrow{ }^{5} C_{0}(x)^{5-0}+{ }^{5} C_{1}(x)^{5-1}\left(-\frac{1}{y}\right)^{1}+{...
Read More →Using binomial theorem, expand each of the following:
Question: Using binomial theorem, expand each of the following: $\left(x^{2}-\frac{3}{x}\right)^{7}$ Solution: To find: Expansion of $\left(x^{2}-\frac{3 x}{7}\right)^{7}$ Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$ (ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$ We have, $\left(x^{2}-\frac{3 x}{7}\right)^{7}$ $\Rightarrow\left[{ }^{7} C_{0}\left(x^{2}\right)^{7-0}\right]+\left[{ }...
Read More →Using binomial theorem, expand each of the following:
Question: Using binomial theorem, expand each of the following: $\left(\frac{2 x}{3}-\frac{3}{2 x}\right)^{6}$ Solution: To find: Expansion of $\left(\frac{2 x}{3}-\frac{3}{2 x}\right)^{6}$ Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$ (ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$ We have, $\left(\frac{2 x}{3}-\frac{3}{2 x}\right)^{6}$ $\Rightarrow\left[{ }^{6} C_{0}\left(\frac{2 x...
Read More →Using binomial theorem, expand each of the following:
Question: Using binomial theorem, expand each of the following: $(2 x-3 y)^{4}$ Solution: To find: Expansion of $(2 x-3 y)^{4}$ Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$ (ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$ We have, $(2 x-3 y)^{4}$ $\Rightarrow\left[{ }^{4} C_{0}(2 x)^{4-0}\right]+\left[{ }^{4} C_{1}(2 x)^{4-1}(-3 y)^{1}\right]+\left[{ }^{4} C_{2}(2 x)^{4-2}(-3 y)^{2}\...
Read More →Using binomial theorem, expand each of the following:
Question: Using binomial theorem, expand each of the following: $(3 x+2 y)^{5}$ Solution: To find: Expansion of $(3 x+2 y)^{5}$ Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$ (ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$ We have, $(3 x+2 y)^{5}$ $\Rightarrow\left[{ }^{5} C_{0}(3 x)^{5-0}\right]+\left[{ }^{5} C_{1}(3 x)^{5-1}(2 y)^{1}\right]+\left[{ }^{5} C_{2}(3 x)^{5-2}(2 y)^{2}\ri...
Read More →Using binomial theorem, expand each of the following:
Question: Using binomial theorem, expand each of the following: $(2 x-3)^{6}$ Solution: To find: Expansion of $(2 x-3)^{6}$ Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$ (ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$ We have, $(2 x-3)^{6}$ $\Rightarrow\left[{ }^{6} \mathrm{C}_{0}(2 \mathrm{x})^{6}\right]+\left[{ }^{6} \mathrm{C}_{1}(2 \mathrm{x})^{6-1}(-3)^{1}\right]+\left[{ }^{6} \...
Read More →Using binomial theorem, expand each of the following:
Question: Using binomial theorem, expand each of the following: $(1-2 x)^{5}$ Solution: To find: Expansion of $(1-2 x)^{5}$ Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$ (ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$ We have, $(1-2 x)^{5}$ $\Rightarrow\left[{ }^{5} \mathrm{C}_{0}(1)^{5}\right]+\left[{ }^{5} \mathrm{C}_{1}(1)^{5-1}(-2 \mathrm{x})^{1}\right]+\left[{ }^{5} \mathrm{C}_{...
Read More →How many parallelograms can be formed from a set of 4 parallel lines
Question: How many parallelograms can be formed from a set of 4 parallel lines interesting another set of 3 parallel lines? Solution: To form a parallelogram we need 2 sets of 2 parallel lines intersecting the other 2 lines from the other set. So, first of all, we need to get 2 lines from the sets. From the first parallel set, 2 out of 4 lines can be selected in ${ }^{4} \mathrm{C}_{2}=6$ ways. From the second parallel set, 2 out of 3 lines can be selected in ${ }^{3} \mathrm{C}_{2}=3$ ways. So,...
Read More →How many different committees of 5 can be formed from 6 men and 4
Question: How many different committees of 5 can be formed from 6 men and 4 women, if each committee consists of 3 men and 2 women? Solution: Each committee consists of 3 men and 2 women. So, we need to select 3 men out of 6 and 2 women out of 4 . The number of ways, 3 men can be selected out of 6 , is $={ }^{6} C_{3}=20$ The number of ways, 2 women can be selected out of 4 , is $={ }^{4} C_{2}=6$ So, the totally $(20+6)=26$ numbers of different committees can be formed....
Read More →There are 13 cricket players, out of which 4 are bowlers
Question: There are 13 cricket players, out of which 4 are bowlers. In how many ways can team of 11 be selected from them so as to include at least 3 bowlers? Solution: There are 4 bowlers in 13 player team. So, maximum we can add 4 bowlers. And we need to include at least 3 bowlers. If we include 3 bowlers then from the remaining 9 [13 - 4 bowlers] players, we need to include 8 . The number of ways, 8 players can be selected among 9 is $={ }^{9} \mathrm{C}_{8}=9$ The number of ways, 3 players c...
Read More →In how many ways can committee of 5 be made out of 6 men
Question: In how many ways can committee of 5 be made out of 6 men and 4 women, containing at least 2 women? Solution: We need to include at least 2 women. If we include 2 women in the committee, then a number of men is 3. The number of ways, 2 women can be selected out of 4 is $={ }^{4} \mathrm{C}_{2}=6$ The number of ways, 3 men can be selected out of 6 is $={ }^{6} \mathrm{C}_{3}=$ 20 So, the committee can be formed including 2 women in $(20 \times 6)=120$ ways. If we include 3 women in the c...
Read More →There are 12 points in a plane, out of which 3 points are collinear.
Question: There are 12 points in a plane, out of which 3 points are collinear. How many straight lines can be drawn by joining any two of them? Solution: To get a straight line we just need to join two points. There are 12 numbers of points. Therefore, there is ${ }^{12} \mathrm{C}_{2}=66$ number of straight lines. Among the 12 points, there are 3 points which are collinear. That means joining those 3 lines give a single straight line. That means the real number of straight lines present in the ...
Read More →Three persons enter a railway compartment having 5 vacant seats.
Question: Three persons enter a railway compartment having 5 vacant seats. In how many ways can they seat themselves? Solution: Three persons enter a compartment where 5 seats are vacant. The number of ways they can be seated is $={ }^{5} \mathrm{P}_{3}=60$...
Read More →Find the number of diagonals in an n-sided polygon.
Question: Find the number of diagonals in an n-sided polygon. Solution: n-sided polygon has n numbers of vertices. Diagonals are formed by joining the opposite vertices from one vertex, except the two adjacent vertices. So, from one vertex $(n-3)$ diagonals can be drawn. Similarly, for $n$ numbers of vertices, $n(n-3)$ diagonals can be drawn. But, the diagonal joins 2 points at a time, here two vertices. Therefore, the actual number of diagonals is $=\frac{n(n-1)}{2}$....
Read More →How many words are formed by 2 vowels and 3 consonants, taken from 4 vowels and 5 consonants?
Question: How many words are formed by 2 vowels and 3 consonants, taken from 4 vowels and 5 consonants? Solution: 3 consonants out of 5 consonants can be chosen in ${ }^{5} C_{3}$ ways. 2 vowels out of 4 vowels can be chosen in ${ }^{4} \mathrm{C}_{2}$ ways. And also 5 letters can be written in $5 !$ Ways. Therefore, the number of words can be formed is $\left({ }^{5} \mathrm{C}_{3} \mathrm{X}^{4} \mathrm{C}_{2} \times 5 !\right)=7200$....
Read More →Solve this
Question: If ${ }^{(n 2-n)} C_{2}={ }^{(n 2-n)} C_{4}=120$ then find the value of $n .$ Solution: Given: ${ }^{(n 2-n)} \mathrm{C}_{2}={ }^{(n 2-n)} \mathrm{C}_{4}=120$ Need to find: Value of $n{ }^{(n 2-n)} \mathrm{C}_{2}={ }^{(n 2-n)} \mathrm{C}_{4}=$ 120 We know, one of the property of combination is: If ${ }^{n} C_{r}={ }^{n} C_{t}$, then, (i) $r=t$ OR (ii) $r+$ $t=n$ Applying property (ii) we get, $n^{2}-n=2+4=6 n^{2}-n-6=0 n^{2}-3 n+2 n-6=$ $0 n(n-3)+2(n-3)=0(n-3)(n+2)=0$ So, the value of ...
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Question: If ${ }^{n} P_{r}=720$ and ${ }^{n} C_{r}=120$ then find the value of $r$. Solution: Given: ${ }^{n} P_{r}=720 \{ }^{n} C_{r}=120$ Need to find: Value of $r$ We know that, ${ }^{n} P_{r}=r ! X{ }^{n} C_{r}$ Putting the values, $\Rightarrow 720=r ! X 120 \Rightarrow r !=6=3 ! \Rightarrow r=3$....
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Question: If ${ }^{n+1} C_{3}=2\left({ }^{n} C_{2}\right)$, find the value of $n .$ Solution: Given: ${ }^{n+1} C_{3}=2\left({ }^{n} C_{2}\right)$ Need to find: Value of $n \Rightarrow{ }^{n+1} C_{3}=$...
Read More →Write the value of
Question: Write the value of $\left({ }^{5} C_{1}+{ }^{5} C_{2}+{ }^{5} C_{3}+{ }^{5} C_{4}+{ }^{5} C_{5}\right)$ Solution: $\Rightarrow{ }^{5} \mathrm{C}_{1}+{ }^{5} \mathrm{C}_{2}+{ }^{5} \mathrm{C}_{3}+{ }^{5} \mathrm{C}_{4}+{ }^{5} \mathrm{C}_{5} \Rightarrow{ }^{6} \mathrm{C}_{2}+{ }^{6} \mathrm{C}_{4}+1\left[\right.$ As $\left.{ }^{5} \mathrm{C}_{5}=1\right] \Rightarrow 15+15+1 \Rightarrow 31$...
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