If C r - n C r- then C 0 C 4- C 1 C 3- C 2 C 2- C 3 C 1- C 4 C 0-A- C 2B- 2 n Cn-1C- C 22D- 2 n C r

The correct option is A C_2We know that, (1+x)^n=C_0+C_1x+C_2x^2+C_3x^3+⋯+C_nx^n ⋯(1) (1 x)^n=C_0 C_1x+C_2x^2 C_3x^3+⋯+( 1)^nC_nx^n ⋯(2) Multiplying (1) a ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Sum of Binomial Coefficients with Alternate Signs

If C r = n C ...

Question

If $C_{r} =^{n} C_{r}$ , then $C_{0} C_{4} - C_{1} C_{3} + C_{2} C_{2} - C_{3} C_{1} + C_{4} C_{0} =$

(C_{2})^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

^{2 n} C_{r}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C_{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

^{2 n} C_{n - 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $C_{2}$
We know that,
$(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + C_{3} x^{3} + \dots + C_{n} x^{n} \dots (1)$
$(1 - x)^{n} = C_{0} - C_{1} x + C_{2} x^{2} - C_{3} x^{3} + \dots + (- 1)^{n} C_{n} x^{n} \dots (2)$

Multiplying $(1)$ and $(2)$ and equating the coefficients of $x^{4},$ we get
$C_{0} C_{4} - C_{1} C_{3} + C_{2} C_{2} - C_{3} C_{1} + C_{4} C_{0} = C_{2} (- 1)^{2} = C_{2}$

Suggest Corrections

Similar questions

Find the coefficient of $x^{4}$ in the expansion of $(1 + x)^{n} (1 - x)^{n}$ . Deduce that $C_{2} = C_{0} C_{4} - C_{1} C_{3} + C_{2} C_{2} - C_{3} C_{1} + C_{4} C_{0},$ where, $C$ stands for $^{n} C_{r} .$