સરવાળો શોધો: $\left( \binom{21}{1} - \binom{10}{1} \right) + \left( \binom{21}{2} - \binom{10}{2} \right) + \left( \binom{21}{3} - \binom{10}{3} \right) + \dots + \left( \binom{21}{10} - \binom{10}{10} \right) = $

  • A
    $2^{20} - 2^{10}$
  • B
    $2^{21} - 2^{11}$
  • C
    $2^{21} - 2^{10}$
  • D
    $2^{20} - 2^9$

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જો $\frac{1}{n+1} {}^{n}C_{n} + \frac{1}{n} {}^{n}C_{n-1} + \dots + \frac{1}{2} {}^{n}C_{1} + {}^{n}C_{0} = \frac{1023}{10}$ હોય,તો $n$ ની કિંમત શોધો.

જો $^nC_r = C_r$ અને $2 \frac{C_1}{C_0} + 4 \frac{C_2}{C_1} + 6 \frac{C_3}{C_2} + \dots + 2n \frac{C_n}{C_{n-1}} = 650$ હોય,તો $^nC_2 =$

ધારો કે $\binom{n}{k}$ એ ${}^{n}C_{k}$ દર્શાવે છે અને $\left[\begin{array}{c} n \\ k \end{array}\right]=\begin{cases} \binom{n}{k}, & \text{જો } 0 \leq k \leq n \\ 0, & \text{અન્યથા} \end{cases}$. જો $A_{k}=\sum_{i=0}^{9}\binom{9}{i}\left[\begin{array}{c} 12 \\ 12-k+i \end{array}\right]+\sum_{i=0}^{8}\binom{8}{i}\left[\begin{array}{c} 13 \\ 13-k+i \end{array}\right]$ અને $A_{4}-A_{3}=190p$ હોય,તો $p$ ની કિંમત શોધો:

ધારો કે $S_1 = \sum_{j=1}^{10} j(j-1) \binom{10}{j}$,$S_2 = \sum_{j=1}^{10} j \binom{10}{j}$,અને $S_3 = \sum_{j=1}^{10} j^2 \binom{10}{j}$.
વિધાન $(A) : S_3 = 55 \times 2^9$
કારણ $(R) : S_1 = 90 \times 2^8$ અને $S_2 = 10 \times 2^8$

વિસ્તરણ $(1+x+x^2)^n = a_0 + a_1 x + a_2 x^2 + \ldots + a_{2n} x^{2n}$ માટે યાદી-$I$ માં આપેલા પદોને યાદી-$II$ માં આપેલા તેમના મૂલ્યો સાથે જોડો.
યાદી-$I$યાદી-$II$
$(A)$ $a_0 + a_2 + \ldots + a_{2n}$$(I)$ $n \cdot 3^{n-1}$
$(B)$ $a_1 + a_3 + \ldots + a_{2n-1}$$(II)$ $n \cdot 3^n$
$(C)$ $a_1 + 2a_2 + 3a_3 + \ldots + 2n a_{2n}$$(III)$ $\frac{1}{2}(3^n + 1)$
$(IV)$ $\frac{1}{2}(3^n - 1)$

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