JEE MAIN PT-02 Mathematics
Exam Duration: 60 Mins Total Questions : 25
For \(1\le r\le n\),the value of \(^{ n }{ C }_{ r }+^{ n-1 }{ C }_{ r }+^{ n-2 }{ C }_{ r }+......^{ r }{ C }_{ r },\)
- (a)
\(_{ }^{ n }{ C }_{ r+1 }\)
- (b)
\(_{ }^{ n +1}{ C }_{ r }\)
- (c)
\(_{ }^{ n+1 }{ C }_{ r+1 }\)
- (d)
None of these
Let S='C,+r+1c,+...+n-2C,+n-1C,+nC, ..(i)
(writing the terms in reverse order)
We know that the expansion of (1+x)n , coeffiecient of
xr+1 is nC
Consider the series
S1=(1+x)r+(1+x)r+1+....+(1+x)n-2+(1+x)n-1+(1+x)n ...(ii)
The series R.H.S of (ii) is G.P with common ratio (1+x) having (n+1)-r terms
∴ \({ S }_{ 1 }=\frac { (1+x)\left[ { (1+x) }^{ n-r+1 }-1 \right] }{ (1+x-1) } \)
= \(\frac { { (1+x) }^{ n+1 }-{ (1+x) }^{ r } }{ x } \) ...(iii)
Coefficient of xr+1 = Coefficient of xr+2 in the expansion of (1+x)n+1-(1+x)r
=n+1Cr+1-0=n+1Cr+1
∴ nCr +n-1Cr+n-2Cr+....rCr=n+1Cr+1
If x=9950+10050 and y=10150,then
- (a)
x=y
- (b)
x<y
- (c)
x>y
- (d)
None of these
Coefficient of x99 in the polynomial (x-1)(x-2)(x-3)....(x-100) is
- (a)
1000
- (b)
1002
- (c)
-5050
- (d)
NONE OF THESE
We know,
(x-a)(x-b)(x-c)=x3-(a+b+c)x2+(ab+bc+ca)x2-(ab+bc+ca)x-abc
Thus, sum of coefficients of x2 is -(a+b+c).
Hence, sum of coefficients of x99 in (x-1)(x-2)(x-3)...(x-100) is
-(1+2+3...+100)\(=-\frac { 100\times (100+1) }{ 2 } \)
=-50x101=-5050.
The sum of integers from 1 to 100 that are divisible by 2 or 5 is
- (a)
2550
- (b)
1050
- (c)
550
- (d)
3050
The required sum
=(2+4+6+8+10+12+...+100)+(5+15+25...95)
=2(1+2+3+4+5+6+...+50)+5(1+3+5+...+19)
\(=2\times \frac { 50\times \left( 50+1 \right) }{ 2 } +5\times \frac { 10 }{ 2 } \left[ 2+(10-1)\times 2 \right] \)
=2550+500=3050
Value of \(1+log\quad y+\frac { { (log\quad y })^{ 2 } }{ 2! } +\frac { { (log\quad y) }^{ 3 } }{ 3! } +...\infty \) is
- (a)
log y
- (b)
- log y
- (c)
2 log y
- (d)
y
Given series=elog y=y
For every positive integer values of n, 32n-2n+1 is divisible by
- (a)
12
- (b)
4
- (c)
8
- (d)
2
On putting n=2, we get \({ 3 }^{ 2\times 2 }-2\times 2+1={ 3 }^{ 4 }-4+1=81-3=78\)
So, p(n) is divisible by 2.
For all n ϵ N, which one of the following is true?
- (a)
\(cos\theta .cos2\theta .cos4\theta .\quad .....\quad .\quad cos(2^{ n-1 }\theta )=\frac { sin{ 2 }^{ n }\quad \theta }{ 2^{ n }sin\quad \theta } \)
- (b)
\(sin\theta .sin2\theta .sin4\theta .\quad .....\quad .\quad sin(2^{ n-1 }\theta )=\frac { sin{ 2 }^{ n }\quad \theta }{ 2^{ n }sin\quad \theta } \)
- (c)
\(sin\theta .cos2\theta .sin4\theta .\quad .....\quad .\quad cos(2^{ n-1 }\theta )=\frac { sin{ 2 }^{ n }\quad \theta }{ 2^{ n }sin\quad \theta } \)
- (d)
None of the above
Let P(n): \(cos\theta
.cos2\theta.cos4\theta.....cos(2^{n-1}\theta)={sin2^n\theta\over 2^nsin\theta}\)
\(∴\ \ P(1):cos\theta={sin2\theta\over 2sin\theta}=cos\theta\)
\(P(2):cos\theta.cos2\theta={sin4\theta\over 4sin\theta}=-{2sin2\theta.cos2\theta\over 4sin\theta}\)
\(={2.2.sin\theta.cos\theta.cos2\theta\over 4sin\theta}\)
=cosθ.cos2θ
So, we have to prove the option () by mathematical induction for P(n)
Step I P(1): \(cosθ={sin2θ\over 2sinθ}=cosθ\)
So, P(1) is true
Step II Let P(m) be true
\(P(m): cosθ.cos2θ.cos4θ....cos(2^{m-1}θ).{sin(2^mθ)\over 2^msinθ}\)
where, m∈N
Step III
\(P(m): cosθ.cos2θ.cos4θ....cos(2^{m-1}θ).cos(2^mθ)\)
\(={sin2.2^mθ\over 2^{m+1}sin θ}={sin^2{m+1}θ\over 2^{m+1}sinθ}\)
So, P(m+1) is true
Hence, P(n) is true for all natural number
For all positive integers n > 1, 23n-7n-1 is divisible by
- (a)
49
- (b)
47
- (c)
13
- (d)
None of these
Let P(n) \(\equiv \) 23n-7n-1
\(\therefore \) P(1) \(\equiv \) 0, P(2) = 49
P(1) and P(2) are divisible by 49.
Let P(k) \(\equiv \) 23k-7k-1=49\(\lambda \)
\(\therefore \) P(k+1) \(\equiv \) 23k+3-7k-8=8(49\(\lambda \)+7k+1)-7k-8=49(8\(\lambda \))+49k=49\(\mu\)
Hence, by mathematical induction, 23n-7n-1 is divisible by 49.
If P(n): 49n+16n+\(\lambda \) is divisible by 64 for n ϵ N, then the least negative integral value of \(\lambda \) is
- (a)
-1
- (b)
-2
- (c)
-3
- (d)
-4
Let P(n) : 49n+16n+\(\lambda \)
\(\therefore \) P(1) : 49+16+\(\lambda \)=65+\(\lambda \)
So, the least value of \(\lambda \) is -1.
The number of terms in the expansion of \(\left( 1+5\sqrt { 2x } \right) ^{ 19 }+\left( 1-5\sqrt { 2x } \right) ^{ 19 }\) is
- (a)
10
- (b)
7
- (c)
13
- (d)
4
\(\left( 1+5\sqrt { 2x } \right) ^{ 19 }+\left( 1-5\sqrt { 2x } \right) ^{ 19 }\)
\(=2\left[ ^{ 19 }{ C }_{ 0 }+^{ 19 }{ C }_{ 2 }\left( 5\sqrt { 2x } \right) ^{ 2 }+....+^{ 19 }{ C }_{ 18 }\left( 5\sqrt { 2x } \right) ^{ 18 } \right] \)
So, it has only 10 terms.
If the ratio of the term from the beginning to the fifth term from the end in the expansion of \(\left( \sqrt [ 4 ]{ 2 } +\frac { 1 }{ \sqrt [ 4 ]{ 3 } } \right) ^{ n }\) is \(\sqrt { 6 } :1\), then the value of n is
- (a)
7
- (b)
8
- (c)
9
- (d)
10
\(\left(\sqrt[4]{2}+{1\over \sqrt[4]{3}}\right)^n=\left(2^{1/4}+{1\over3^{1/4}}\right)^n=(2^{1/4}+3^{-1/4})^n\)
Now, from beginning, T4+1=T5=nC4(21/4)n-4(21/4)4
\(T_5= ^4C_42^{n-4\over 4}3^{-1}\) ...(i)
From end, \(\left( 3^{1\over 4}+2^{1\over 4}\right)^n, T_5=T_{4+1}= ^nC_4(3^{-1/4})^{n-4}(2^{1/4})^4\)
\(∴\ \ T_5= ^nC_43^{-{n-4\over 4}}.2^1\) ....(ii)
According to the given condition, \({Fifth\ term\ from\ the\ beginning\over Fifth\ term\ from\ the\ end}={\sqrt6\over 1}\)
\(∴\ \ {^nC_42^{n-4\over 4}3^{-1}\over ^nC_43^{-{n-4\over 4}}2^{-1}}={\sqrt6\over 1}⇒2^{{n-8\over 4}-1}3^{3^{-1}{n-8\over 4}}=\sqrt6\)
\(⇒\ 2^{n-4-4\over 4}3^{-4+n-4\over 4}={\sqrt6}⇒2^{n-8\over 4}3^{n-8\over 4}=\sqrt6\)
\(⇒\ (6)^{4-8\over 8}=6^{1/2}\)
On comparing power of 6, we get
\({n-8\over 4}={1\over 2}⇒n-8={4\over 2}\)
⇒ n-8=2 ⇒ n = 10
The value of\(\overset { n }{ \underset { r=0 }{ \Sigma } } { r^{ 2 } }\quad ^{ n }{ C_{ r }{ x }^{ r } }.{ y }^{ n-r },\) when x+y=1, is
- (a)
nx(x+y)
- (b)
nx(nx+y)
- (c)
nx(x-y)
- (d)
nx(nx-y)
\(\sum_{r=0}^n\ ^nC_r x^r y^{n-r}=\sum_{r=0}^n[r(r-1)+r]\ ^nC_rx^ry^{n-r}\)
\(=\sum_{r=0}^nr(r-1)\ {n\over r}\ ^nC_rx^ry^{n-r}+\sum_{r=0}^nr \ ^nC_rx^ry^{n-r}\)
\(=\sum_{r=2}^{n-2}r(r-1){n\over r}.{n-1\over r-1}\ ^{n-2}C_{r-2}x^2x^{r-2}y^{n-r}+\sum_{r=1}^{n-1}r{n\over r}\ ^{n-1}C_{r-1}xx^{r-1}.y^{n-r}\)
\(=n(n-1)x^2\sum_{r=2}^{n-2}\ ^{n-2}C_{r-2}x^{r-2}y^{(n-2)-(r-2)}+nx\sum_{r=1}^{n-1}\ ^{n-1}C_{r-1}x^{r-1}y^{(n-1)-(r-1)}\)
=n(n-1)x2(x+y)n-2+nx(x+y)n-1
=n(n-1)x2+nx [∵ x+y=1]
=nx(nx-x+1)=nx(nx+y) [∵ x+y=1]
The sum upto n terms of the sequence log a, log ar, log ar 2 , .... is
- (a)
\(\frac { n }{ 2 } \log { { a }^{ 2 }{ r }^{ n-1 } } \)
- (b)
\(\frac { n }{ 2 } \log { { a }{ r }^{ n-1 } } \)
- (c)
\(\frac { 3n }{ 2 } \log { { a }{ r }^{ n-1 } } \)
- (d)
\(\frac { 5n }{ 2 } \log { { a }^{ 2 }{ r }^{ n-1 } } \)
The sequence can be written as log a, (log a + log r), (log a + 2log r), .....This is an AP, where A = log a, D = logr
Now, \({ T }_{ n }=A+\left( n-1 \right) D=\log { a } +\left( n-1 \right) \log { r } =1\)
\(\therefore \quad { S }_{ n }=\frac { n }{ 2 } \left( A+I \right) =\frac { n }{ 2 } \left[ \log { a } +\log { a } +\left( n-1 \right) \log { r } \right] \)
\(=\frac { n }{ 2 } \left[ \log { { a }^{ 2 }+\log { { r }^{ n-1 } } } \right] =\frac { n }{ 2 } \log { { a }^{ 2 }{ r }^{ n-1 } } \)
Statement I The sum of the series 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + .. + (361 + 380 + 400) is 8000.
Statement II \(\sum _{ k=1 }^{ n }{ \left[ \left( { k }^{ 3 }-\left( { k-1 }^{ 3 } \right) \right) \right] } ={ n }^{ 3 },\) for any natural number n.
- (a)
Statement I is true, Statement II is true, Statement II is the correct explanation for statement I
- (b)
Statement I is true, Statement II is true, Statement II is not the correct explanation for statement I
- (c)
Statement I is true, Statement II is false
- (d)
Statement I is false, Statement II is true.
Statement I
S=(1)+(1+2+4)+(4+6+9)+(9+12+16)+...+(361+380+400)
S=(0+0+1)+(1+2+4)+(4+6+9)+(9+12+16)+...+(361+380+400)
Now, we can clearly, observe the first elements in each bracket.
In second bracket, the first element is 1 = 12
In third bracket, the first element is 4 = 22
In fourth bracket, the first element is 9 = 32
In last bracket, the first element is 361 = 192
Hence, we can conclude that there are 20 brackets in all. Also, in each of the brackets there are 3 terms out of which the first and last terms are perfect squares of consecutive integers and the middle term is their product.
∴ The general term of the series is
Tr=(r-1)2+(r+1)r+(r)2
Now, the sum of the n terms of the series is
\({ S }_{ n }=\sum _{ r=1 }^{ n }{ \left[ \left( r-1 \right) ^{ 2 }+(r-1)r+(r)^{ 2 } \right] } \)
\({ S }_{ n }=\sum _{ r=1 }^{ n }{ \left\{ \frac { { r }^{ 3 }-\left( r-1 \right) ^{ 3 } }{ r-(r-1) } \right\} } \)
\(\left[ \because \quad { a }^{ 3 }-{ b }^{ 3 }=\left( a-b \right) \left( { a }^{ 2 }+ab+{ b }^{ 2 } \right) \right] \)
\(\Rightarrow { S }_{ n }=\sum _{ r=1 }^{ n }{ \left[ { r }^{ 3 }-(r-1)^{ 3 } \right] } \)
\({ S }_{ n }=\left( { 1 }^{ 3 }-{ 0 }^{ 3 } \right) +\left( { 2 }^{ 3 }-{ 1 }^{ 3 } \right) +\left( { 3 }^{ 3 }-{ 2 }^{ 3 } \right) +...+\left\{ { n }^{ 3 }-\left( n-1 \right) ^{ 3 } \right\} \)
On rearranging the terms, we get
\({ S }_{ n }=-{ 0 }^{ 3 }+\left( { 1 }^{ 3 }-{ 1 }^{ 3 } \right) +\left( { 2 }^{ 3 }-{ 2 }^{ 3 } \right) +\left( { 3 }^{ 3 }-{ 3 }^{ 3 } \right) +...+\left[ \left( n-1 \right) ^{ 3 }-\left( n-1 \right) ^{ 3 } \right] +{ n }^{ 3 }\)
\(\Rightarrow { S }_{ n }={ n }^{ 3 }\)
Since, the number of terms is 20, hence on substituting n = 20
we get S20 = 8000
Hence, Statement I is correct.
Statement II We have already proved in the Statement l, that Statement II is true.
If a3 + b6 =2, then the maximum value of the term independent of x in the expansion of (ax1/3 + bx -1/6)9 is (a>0, b>0)
- (a)
42
- (b)
68
- (c)
84
- (d)
148
Let (r+1) th term be independent of x, then
\({ T }_{ \left( r+1 \right) }=^{ 9 }{ { C }_{ r } }\left( { ax }^{ \frac { 1 }{ 3 } } \right) \left( { bx }^{ -1/6 } \right) ^{ r }\)
\(=^{ 9 }{ { C }_{ r } }{ a }^{ 9-r }.{ b }^{ r }{ x }^{ 3-\frac { r }{ 3 } -\frac { r }{ 6 } }\)
As Tr+1 is independent of x
\(\therefore \quad \quad \quad \quad \quad 3-\frac { r }{ 3 } -\frac { r }{ 6 } =0\)
\(\Rightarrow \quad \quad \quad \quad \quad \quad \quad \quad \quad r=6\)
\(\therefore \quad \quad { T }_{ 6+1 }=^{ 9 }{ { C }_{ 6 } }{ a }^{ 3 }{ b }^{ 6 }=^{ 9 }{ { C }_{ 3 } }{ a }^{ 3 }{ b }^{ 6 }=84{ a }^{ 3 }{ b }^{ 6 }\)
\(\therefore \quad \quad \quad \quad \quad \quad \quad \quad AM\ge GM\)
\(\therefore \quad \quad \quad \quad \quad \frac { { a }^{ 3 }+{ b }^{ 6 } }{ 2 } \ge \sqrt { { a }^{ 3 }{ b }^{ 6 } } \)
\(\Rightarrow \quad \quad \quad \quad \quad \frac { 2 }{ 2 } \ge \sqrt { { a }^{ 3 }{ b }^{ 6 } } \)
\(\Rightarrow \quad \quad \quad \quad \quad \quad \quad { a }^{ 3 }{ b }^{ 6 }\le 1\)
or \(84{ a }^{ 3 }{ b }^{ 6 }\le 84\)
or \({ T }_{ 7 }\le 84\)
Maximum value = 84
(115)96 - (96)115 is divided by
- (a)
15
- (b)
17
- (c)
19
- (d)
21
(115)96 - (96)115 = (1 + 114)96 - (1 + 95)115
= {1 + 96C1 (114) + 96C2 (114)2 +....} - {1 + 115C1 (95) + 115C2 (95)2 +.....}
= 114 {96C1 + 96C2 (114) + ....} -95 {115C1 + 115C2 (95) +......}
= 19 {6 (96C1 + 96C2 (95) +......) - 5 (115C1 + 115C2 (95) +......) }
Hence, given expression is divisible by 19.
If A, G and H are respectively arithmetic, geometric and harmonic means between a and b both being unequal and positive. then \(A=\frac { a+b }{ 2 } \Rightarrow a+b=2A,G=\sqrt { ab } \Rightarrow ab={ G }^{ 2 }and\quad H=\frac { 2ab }{ a+b } \Rightarrow { G }^{ 2 }=AH\)
From above discussion we can say that a. b are the roots of the equation x2-- 2A x + G2 = 0
Now, quadratic equation x2--Px + Q=0 and quadratic equation a (b - c)x2 + b (c - a)x + c (a - b) = 0 have a root common and satisfy the relation b = \(\frac { 2ac }{ (a+c) } \),where a, b,c are real numbers.
If the geometric and harmonic means of two numbers are 16 and 12 \(\frac { 4 }{ 5 } \) , then the ratio of one number to the other is
- (a)
1 : 4
- (b)
2 : 3
- (c)
1 : 2
- (d)
2 : 1
\(\sqrt { ab } =16\)
\(\Rightarrow \quad ab={ 16 }^{ 2 }\)
\(\Rightarrow\) ab = 8.32
and \(\frac { 2ab }{ a+b } =12\frac { 4 }{ 5 } \)
\(\Rightarrow \frac { 2\times 256 }{ a+b } =\frac { 64 }{ 5 } \)
\(\Rightarrow \quad a+b=40=8+32\)
\(\therefore \quad \frac { a }{ b } =\frac { 8 }{ 32 } =\frac { 1 }{ 4 } \)
\(\Rightarrow \) a : b = 1 : 4
Using principle of mathematical induction,
- (a)
9 ∀n∈N
- (b)
11 ∀n∈N
- (c)
13 ∀n∈N
- (d)
15 ∀n∈N
Let P(n) be the statement given
P(n) :4n+15n-1 is divisible by 9
For n=1,P(1):41+15x1-1=18, which is divisible by 9
∴ P(1) is true.
Let P(k) be true. Then,
4k+15k-1is divisible by 9
⇒4k+15k-1 =9λ, for some λ∊N.
We shall now show that P(k+1) is true, for thisa we have to show that 4k+1+15(k+1)-1 is divisible by 9.
Now, 4k+1 +15(k+1)-1
=4k.4+15(k+1)-1
=(9λ-15k+1)x4+15(k+1)-1
=36λ-45k+18
=9(4λ-5k+2), which is divisible by 9
∴P(k+1) is true.
Thus, P(k) is true ⇒P(k+1) is true.
Hence, by the principle of mathematical induction P(n) is true for all n ∈N.
If m,n are any two odd positive integers with n<m, then the largest positive integers which divides all the numbers of the type m2-n2 is
- (a)
4
- (b)
6
- (c)
8
- (d)
9
Let m=2k+1, n=2k-1(k∈N)
∴m2-n2=4k2+1+4k-4k2+4k-1=8k
Hence, all the numbers of the form m2-n2 are always divisible by 8
23n-7n-1 is divisible by
- (a)
64
- (b)
36
- (c)
49
- (d)
25
Let P(n) =23n -7n-1 ∴ P(1)=0, P(2)=49
P(1) and P(2)are divisible by 49.
Let P(k)☰23k-7k -1=49λ
∴ P(k+1)☰23k+3-7(k+1)-1 ...(i)
=8(49λ+7k+1)-7k-8 [using(i)]
=49(8λ)+49k=49(8λ+k)
Hence, by the principle of mathematical induction 23n-7n-1 is divisible by 49
If xn-1 is divisible by x-k, then the least positive integral value of k is
- (a)
1
- (b)
2
- (c)
3
- (d)
4
x-1 is always a factor of xn-1. ⇒k=1
The coefficient of x7 in the expansion of \(\left( \frac { { x }^{ 2 } }{ 2 } -\frac { 2 }{ x } \right) ^{ 9 }\) is
- (a)
-56
- (b)
14
- (c)
-14
- (d)
None of these
\({ T }_{ r+1 }=^{ 9 }{ C }_{ r }\left( \frac { { x }^{ 2 } }{ 2 } \right) ^{ 9-r }\left( -\frac { 2 }{ x } \right) ^{ 5 }\)
= \(^{ 9 }{ C }_{ r }\left( \frac { 1 }{ 2 } \right) ^{ 9-r }(-2)^{ r }{ x }^{ 18-3r }\)
Now, for the coefficient of x7 , we put 18 - 3r = 7
\(\therefore\) r = 11/3 (not possible)
So, there is no term containing x7
Sum to 10 terms of the series 1+2(1.1)+3(1.1)2+4(1.1)3+..., is
- (a)
85.12
- (b)
92.5
- (c)
96.75
- (d)
None of these
Let x=1.1, then
S=1+2x+3x2+...+10x9.....(1)
Sx=x+2x2+...+9x9+10x10....(2)
Subtracting 2 from 1, we get
s(1-x)=1+x+x2+....+x9-10x10
=\({x^{10}-1\over x-1}-10x^{10}={x^{10}-1\over0.1}-10x^{10},since \ x=1.1\)
=10(x10-1)-10x10=-10
\(\therefore S={10\over x-1}={10\over 0.1}=100\)
The sum of the series 12+(12+22)+(12+22+32)to n terms=...
- (a)
\({n(n+1)^2(n+2)\over12}\)
- (b)
\({n(n+1)^2(n+3)\over12}\)
- (c)
\({n(n+2)^2(n+1)\over12}\)
- (d)
\({n(n+2)^2(n+1)\over14}\)
Tn=(12+22+32+...+n2)
\(={n(n+1)(2n+1)\over6}={1\over6}[2n^3+3n^2+n]\)
\(\therefore S_n=\sum^{n}_{k=1}T_k={1\over6}[2\sum^{n}_{k=1}k^3+3\sum^{n}_{k=1}k^2+\sum^{n}_{k=1}k]\)
\(={1\over6}[2\{{n(n+1)\over 2}\}^2+{3n(n+1)(2n+1)\over6}+{n(n+1)\over2}]\)
\(={1\over6}[{n^2(n+1)^2\over 2}+{n(n+1)(2n+1)\over2}+{n(n+1)\over2}]\)
\(={n(n+1)\over 12}[{n(n+1)+2n+1+1\over 1}]={n(n+1)^2(n+2)\over12}\)
Statement I : The term Independent of x in the expansion of \(\left( x+\frac { 1 }{ x } +2 \right) ^{ m }\) is \(\frac { \left( 4m \right) ! }{ (2m!)^{ 2 } } \)
Statement II : The coefficient of x6 in the expansion of (1 +x)n is nC6
- (a)
If both Statement-I and Statement-II are true and Staternent-Il is the correct explanation of Statement -I.
- (b)
If both Statement-I and Statement-Il are true but Statement-II is not the correct explanation of Statement -I.
- (c)
If Statement-I is true but Statement-II is false.
- (d)
If Statement -I is false and Statement-II is true.
\(\left( x+\frac { 1 }{ x } +2 \right) ^{ m } =\left( \frac { { x }^{ 2 }+2x+1 }{ x } \right) =\frac { \left( 1+x \right) ^{ 2m } }{ { x }^{ m } } \)
Term independent of x is coefficient of xm in the expansion of (1 + x)2m = 2mCm = \(\frac { \left( 2m \right) ! }{ (m!)^{ 2 } } \)
Coefficient of x6 in the expansion of (1 +x)n is nC6
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