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Question: Which of the following is/are correct...

Which of the following is/are correct

A

30C0⋅90C60−30C1⋅89C60+30C288C60+⋯+30C3060C60=60C30{ }^{30}C_0 \cdot { }^{90}C_{60} - { }^{30}C_1 \cdot { }^{89}C_{60} + { }^{30}C_2 { }^{88}C_{60} + \dots + { }^{30}C_{30} { }^{60}C_{60} = { }^{60}C_{30}

B

30C0⋅90C30−30C1⋅89C30+30C288C30+⋯+30C3060C30=0{ }^{30}C_0 \cdot { }^{90}C_{30} - { }^{30}C_1 \cdot { }^{89}C_{30} + { }^{30}C_2 { }^{88}C_{30} + \dots + { }^{30}C_{30} { }^{60}C_{30} = 0

C

30C0⋅90C30−30C1⋅88C30+30C286C30+⋯+30C3030C30=(2)30{ }^{30}C_0 \cdot { }^{90}C_{30} - { }^{30}C_1 \cdot { }^{88}C_{30} + { }^{30}C_2 { }^{86}C_{30} + \dots + { }^{30}C_{30} { }^{30}C_{30} = (2)^{30}

D

30C0⋅90C30−30C1⋅87C30+30C284C30+⋯+30C2030C30=1{ }^{30}C_0 \cdot { }^{90}C_{30} - { }^{30}C_1 \cdot { }^{87}C_{30} + { }^{30}C_2 { }^{84}C_{30} + \dots + { }^{30}C_{20} { }^{30}C_{30} = 1

Answer

A, C

Explanation

Solution

The problem asks us to verify the correctness of four given combinatorial identities. We will analyze each option separately.

Option A:

The given identity is SA=∑r=030(−1)r30Cr90−rC60S_A = \sum_{r=0}^{30} (-1)^r { }^{30}C_r { }^{90-r}C_{60}.
We can express 90−rC60{ }^{90-r}C_{60} as the coefficient of x60x^{60} in the expansion of (1+x)90−r(1+x)^{90-r}.
So, SA=∑r=030(−1)r30Cr[x60](1+x)90−rS_A = \sum_{r=0}^{30} (-1)^r { }^{30}C_r [x^{60}] (1+x)^{90-r}, where [xk]P(x)[x^k] P(x) denotes the coefficient of xkx^k in the polynomial P(x)P(x).
SA=[x60]∑r=030(−1)r30Cr(1+x)90−rS_A = [x^{60}] \sum_{r=0}^{30} (-1)^r { }^{30}C_r (1+x)^{90-r}
SA=[x60](1+x)90∑r=030(−1)r30Cr(1+x)−rS_A = [x^{60}] (1+x)^{90} \sum_{r=0}^{30} (-1)^r { }^{30}C_r (1+x)^{-r}
SA=[x60](1+x)90∑r=03030Cr(−11+x)rS_A = [x^{60}] (1+x)^{90} \sum_{r=0}^{30} { }^{30}C_r \left(-\frac{1}{1+x}\right)^r
Using the binomial theorem ∑r=0nnCryr=(1+y)n\sum_{r=0}^{n} { }^{n}C_r y^r = (1+y)^n, with n=30n=30 and y=−11+xy = -\frac{1}{1+x}, we get:
SA=[x60](1+x)90(1−11+x)30S_A = [x^{60}] (1+x)^{90} \left(1 - \frac{1}{1+x}\right)^{30}
SA=[x60](1+x)90(1+x−11+x)30S_A = [x^{60}] (1+x)^{90} \left(\frac{1+x-1}{1+x}\right)^{30}
SA=[x60](1+x)90(x1+x)30S_A = [x^{60}] (1+x)^{90} \left(\frac{x}{1+x}\right)^{30}
SA=[x60](1+x)90x30(1+x)30S_A = [x^{60}] (1+x)^{90} \frac{x^{30}}{(1+x)^{30}}
SA=[x60]x30(1+x)60S_A = [x^{60}] x^{30} (1+x)^{60}
SA=[x30](1+x)60S_A = [x^{30}] (1+x)^{60}
The coefficient of x30x^{30} in (1+x)60(1+x)^{60} is 60C30{}^{60}C_{30}.
So, the left side of identity A is 60C30{}^{60}C_{30}. The right side is also 60C30{}^{60}C_{30}.
Thus, option A is correct.

Option B:

The given identity is SB=∑r=030(−1)r30Cr90−rC30S_B = \sum_{r=0}^{30} (-1)^r { }^{30}C_r { }^{90-r}C_{30}.
We can express 90−rC30{ }^{90-r}C_{30} as the coefficient of x30x^{30} in the expansion of (1+x)90−r(1+x)^{90-r}.
So, SB=∑r=030(−1)r30Cr[x30](1+x)90−rS_B = \sum_{r=0}^{30} (-1)^r { }^{30}C_r [x^{30}] (1+x)^{90-r}.
SB=[x30]∑r=030(−1)r30Cr(1+x)90−rS_B = [x^{30}] \sum_{r=0}^{30} (-1)^r { }^{30}C_r (1+x)^{90-r}
SB=[x30](1+x)90∑r=030(−1)r30Cr(1+x)−rS_B = [x^{30}] (1+x)^{90} \sum_{r=0}^{30} (-1)^r { }^{30}C_r (1+x)^{-r}
SB=[x30](1+x)90(1−11+x)30S_B = [x^{30}] (1+x)^{90} \left(1 - \frac{1}{1+x}\right)^{30}
SB=[x30](1+x)90(x1+x)30S_B = [x^{30}] (1+x)^{90} \left(\frac{x}{1+x}\right)^{30}
SB=[x30]x30(1+x)60S_B = [x^{30}] x^{30} (1+x)^{60}
SB=[x0](1+x)60S_B = [x^{0}] (1+x)^{60}
The coefficient of x0x^{0} in (1+x)60(1+x)^{60} is 60C0=1{}^{60}C_{0} = 1.
So, the left side of identity B is 1. The right side is 0.
Thus, option B is incorrect.

Option C:

The given identity is SC=30C0⋅90C30−30C1⋅88C30+30C286C30−⋯+30C3030C30S_C = { }^{30}C_0 \cdot { }^{90}C_{30} - { }^{30}C_1 \cdot { }^{88}C_{30} + { }^{30}C_2 { }^{86}C_{30} - \dots + { }^{30}C_{30} { }^{30}C_{30}.
The general term of the sum is (−1)r30Cr90−2rC30(-1)^r { }^{30}C_r { }^{90-2r}C_{30} for r=0,1,…,30r=0, 1, \dots, 30.
So, SC=∑r=030(−1)r30Cr90−2rC30S_C = \sum_{r=0}^{30} (-1)^r { }^{30}C_r { }^{90-2r}C_{30}.
We express 90−2rC30{ }^{90-2r}C_{30} as the coefficient of x30x^{30} in (1+x)90−2r(1+x)^{90-2r}.
SC=∑r=030(−1)r30Cr[x30](1+x)90−2rS_C = \sum_{r=0}^{30} (-1)^r { }^{30}C_r [x^{30}] (1+x)^{90-2r}
SC=[x30]∑r=030(−1)r30Cr(1+x)90−2rS_C = [x^{30}] \sum_{r=0}^{30} (-1)^r { }^{30}C_r (1+x)^{90-2r}
SC=[x30](1+x)90∑r=030(−1)r30Cr(1+x)−2rS_C = [x^{30}] (1+x)^{90} \sum_{r=0}^{30} (-1)^r { }^{30}C_r (1+x)^{-2r}
SC=[x30](1+x)90∑r=03030Cr(−1(1+x)2)rS_C = [x^{30}] (1+x)^{90} \sum_{r=0}^{30} { }^{30}C_r \left(-\frac{1}{(1+x)^2}\right)^r
SC=[x30](1+x)90(1−1(1+x)2)30S_C = [x^{30}] (1+x)^{90} \left(1 - \frac{1}{(1+x)^2}\right)^{30}
SC=[x30](1+x)90((1+x)2−1(1+x)2)30S_C = [x^{30}] (1+x)^{90} \left(\frac{(1+x)^2 - 1}{(1+x)^2}\right)^{30}
SC=[x30](1+x)90(1+2x+x2−1(1+x)2)30S_C = [x^{30}] (1+x)^{90} \left(\frac{1+2x+x^2 - 1}{(1+x)^2}\right)^{30}
SC=[x30](1+x)90(2x+x2(1+x)2)30S_C = [x^{30}] (1+x)^{90} \left(\frac{2x+x^2}{(1+x)^2}\right)^{30}
SC=[x30](1+x)90(x(2+x)(1+x)2)30S_C = [x^{30}] (1+x)^{90} \left(\frac{x(2+x)}{(1+x)^2}\right)^{30}
SC=[x30](1+x)90x30(2+x)30(1+x)60S_C = [x^{30}] (1+x)^{90} \frac{x^{30} (2+x)^{30}}{(1+x)^{60}}
SC=[x30]x30(1+x)30(2+x)30S_C = [x^{30}] x^{30} (1+x)^{30} (2+x)^{30}
SC=[x0](1+x)30(2+x)30S_C = [x^{0}] (1+x)^{30} (2+x)^{30}
SC=[x0]((1+x)(2+x))30S_C = [x^{0}] ((1+x)(2+x))^{30}
SC=[x0](2+3x+x2)30S_C = [x^{0}] (2 + 3x + x^2)^{30}
The constant term (coefficient of x0x^0) is obtained by setting x=0x=0 in the polynomial 2+3x+x22 + 3x + x^2.
So, SC=(2+3⋅0+02)30=230S_C = (2 + 3 \cdot 0 + 0^2)^{30} = 2^{30}.
The left side of identity C is 2302^{30}. The right side is (2)30(2)^{30}.
Thus, option C is correct.

Option D:

The given identity is SD=30C0⋅90C30−30C1⋅87C30+30C284C30−⋯+30C2030C30S_D = { }^{30}C_0 \cdot { }^{90}C_{30} - { }^{30}C_1 \cdot { }^{87}C_{30} + { }^{30}C_2 { }^{84}C_{30} - \dots + { }^{30}C_{20} { }^{30}C_{30}.
The general term of the sum is (−1)r30Cr90−3rC30(-1)^r { }^{30}C_r { }^{90-3r}C_{30} for r=0,1,…,20r=0, 1, \dots, 20.
The last term is for r=20r=20: (−1)2030C2090−3⋅20C30=30C2090−60C30=30C2030C30(-1)^{20} { }^{30}C_{20} { }^{90-3 \cdot 20}C_{30} = { }^{30}C_{20} { }^{90-60}C_{30} = { }^{30}C_{20} { }^{30}C_{30}. This matches the last term in the sum.
So, SD=∑r=020(−1)r30Cr90−3rC30S_D = \sum_{r=0}^{20} (-1)^r { }^{30}C_r { }^{90-3r}C_{30}.
We express 90−3rC30{ }^{90-3r}C_{30} as the coefficient of x30x^{30} in (1+x)90−3r(1+x)^{90-3r}.
SD=∑r=020(−1)r30Cr[x30](1+x)90−3rS_D = \sum_{r=0}^{20} (-1)^r { }^{30}C_r [x^{30}] (1+x)^{90-3r}.
The sum goes up to r=20r=20, not 30. However, for r>20r > 20, 90−3rC30=0{}^{90-3r}C_{30} = 0 because 90−3r<3090-3r < 30 for r>20r > 20. For example, if r=21r=21, 90−3(21)=90−63=27<3090-3(21) = 90-63 = 27 < 30. So 27C30=0{}^{27}C_{30} = 0.
Thus, we can extend the sum up to r=30r=30 without changing the value.
SD=∑r=030(−1)r30Cr90−3rC30S_D = \sum_{r=0}^{30} (-1)^r { }^{30}C_r { }^{90-3r}C_{30}.
SD=[x30]∑r=030(−1)r30Cr(1+x)90−3rS_D = [x^{30}] \sum_{r=0}^{30} (-1)^r { }^{30}C_r (1+x)^{90-3r}
SD=[x30](1+x)90∑r=030(−1)r30Cr(1+x)−3rS_D = [x^{30}] (1+x)^{90} \sum_{r=0}^{30} (-1)^r { }^{30}C_r (1+x)^{-3r}
SD=[x30](1+x)90∑r=03030Cr(−1(1+x)3)rS_D = [x^{30}] (1+x)^{90} \sum_{r=0}^{30} { }^{30}C_r \left(-\frac{1}{(1+x)^3}\right)^r
SD=[x30](1+x)90(1−1(1+x)3)30S_D = [x^{30}] (1+x)^{90} \left(1 - \frac{1}{(1+x)^3}\right)^{30}
SD=[x30](1+x)90((1+x)3−1(1+x)3)30S_D = [x^{30}] (1+x)^{90} \left(\frac{(1+x)^3 - 1}{(1+x)^3}\right)^{30}
SD=[x30](1+x)90(1+3x+3x2+x3−1(1+x)3)30S_D = [x^{30}] (1+x)^{90} \left(\frac{1+3x+3x^2+x^3 - 1}{(1+x)^3}\right)^{30}
SD=[x30](1+x)90(3x+3x2+x3(1+x)3)30S_D = [x^{30}] (1+x)^{90} \left(\frac{3x+3x^2+x^3}{(1+x)^3}\right)^{30}
SD=[x30](1+x)90(x(3+3x+x2)(1+x)3)30S_D = [x^{30}] (1+x)^{90} \left(\frac{x(3+3x+x^2)}{(1+x)^3}\right)^{30}
SD=[x30](1+x)90x30(3+3x+x2)30(1+x)90S_D = [x^{30}] (1+x)^{90} \frac{x^{30} (3+3x+x^2)^{30}}{(1+x)^{90}}
SD=[x30]x30(3+3x+x2)30S_D = [x^{30}] x^{30} (3+3x+x^2)^{30}
SD=[x0](3+3x+x2)30S_D = [x^{0}] (3+3x+x^2)^{30}
The constant term (coefficient of x0x^0) is obtained by setting x=0x=0 in the polynomial 3+3x+x23+3x+x^2.
So, SD=(3+3⋅0+02)30=330S_D = (3 + 3 \cdot 0 + 0^2)^{30} = 3^{30}.
The left side of identity D is 3303^{30}. The right side is 1.
Thus, option D is incorrect.

The correct options are A and C.