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JEE Main 202126 AugustEvening Shift
Exam
JEE Main
Year
2021
Date
26 August
Shift
Evening Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2021 · 26 August · Evening Shift · Mathematics · Binomial Theorem

Let ( array ll n k array ) denote Ck n and [ array ll n k array ] =...

Let \left(nk\begin{array}{ll}n \\ k\end{array}\right) denote CknCk n and \left[nk\begin{array}{ll}n \\ k\end{array}\right] = \left\{nk\begin{array}{ll}n \\ k\end{array}\right, \text{if}0 \le k \le n 0, \text{otherwise}.If. If{A}_{k} = \mathop{∑}_{i = 0}^{9}\left(9i\begin{array}{ll}9 \\ i\end{array}\right)\left[1212−k+i\begin{array}{ll}12 \\ 12 - k + i\end{array}\right] + \mathop{∑}_{i = 0}^{8}\left(8i\begin{array}{ll}8 \\ i\end{array}\right)\left[1313−k+i\begin{array}{ll}13 \\ 13 - k + i\end{array}\right] and A4−A3=190p,{A}_{4} - {A}_{3} = 190p, then pp is equal to _______.

Correct answer

0

Step-by-step solution

Ak=∑i=099Ci⋅Ck−i12+∑i=088Ci⋅Ck−i13{A}_{k} = \mathop{∑}_{i = 0}^{9}{9}_{{C}_{i}} \cdot {C}_{k - i} 12 + \mathop{∑}_{i = 0}^{8}{8}_{{C}_{i}} \cdot {C}_{k - i} 13

=Coefficient of xk{x}^{k} in (1+x)9×(1+x)12+{\left(1 + x\right)}^{9} \times {\left(1 + x\right)}^{12} +Coefficient of xk{x}^{k} in (1+x)8×(1+x)13{\left(1 + x\right)}^{8} \times {\left(1 + x\right)}^{13}

=Coefficient of xk{x}^{k} in (1+x)21+{\left(1 + x\right)}^{21} + Coefficient of xk{x}^{k} in (1+x)21{\left(1 + x\right)}^{21}

Ak=Ck21+Ck21=2.Ck21{A}_{k} = Ck 21 + Ck 21 = 2 . Ck 21

A4=∑i=099Ci⋅C4−i12+∑i=088Ci⋅C4−i13{A}_{4} = \mathop{∑}_{i = 0}^{9}{9}_{{C}_{i}} \cdot {C}_{4 - i} 12 + \mathop{∑}_{i = 0}^{8}{8}_{{C}_{i}} \cdot {C}_{4 - i} 13

=C421+C421=2(C421)= C4 21 + C4 21 = 2\left(C4 21\right)

A3=C321+C321=2(C321){A}_{3} = C3 21 + C3 21 = 2\left(C3 21\right)

A4−A3=2(C421−C321)=190p{A}_{4} - {A}_{3} = 2\left(C4 21 - C3 21\right) = 190p

⇒p=49⇒ p = 49

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