The value of -r-1n nPrr-

The correct option is B 2^n 1 LHS =∑ _ r=1 ^ n ^ n P _ r r! #160; ⇒^ n C _ r = ^ n P _ r r! ∑ _ r=1 ^ n ^ n C _ r = ...

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Standard XII

Mathematics

Combination

The value of ...

Question

The value of $\sum_{r = 1}^{n} \frac{^{n} P_{r}}{r!}$

2^{n}

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2^{n} - 1

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2^{n - 1}

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2^{n + 1}

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Solution

The correct option is B $2^{n} - 1$
$L H S$ $= \sum_{r = 1}^{n} \frac{^{n} P_{r}}{r!}$
$\Rightarrow^{n} C_{r} = \frac{^{n} P_{r}}{r!}$
$\sum_{r = 1}^{n}^{n} C_{r}$
$=^{n} C_{1} +^{n} C_{2} . . . . . . . . .^{n} C_{n}$
$=^{n} C_{0} +^{n} C_{1} +^{n} C_{2} . . . . . . . . .^{n} C_{n} -^{n} C_{0}$
$= 2^{n} - 1$

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Similar questions

Q. If

p, q, r

are in A.P.,then the value of determinant

∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} + 2^{n + 1} + 2 p & b^{2} + 2^{n + 2} + 3 q & c^{2} + p 2^{n} + p & 2^{n + 1} + q & 2 q a^{2} + 2^{n} + p & b^{2} + 2^{n + 1} + 2 q & c^{2} - r \end{matrix} ∣ ∣ ∣ ∣ ∣

Q. If

(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n},

then
the value of

\sum \sum 0 \leq r < s \leq n (r + s) C_{r} C_{s}

The value of $\sum_{r = 1}^{n + 1} (\sum_{k = 1}^{n}^{k} C_{r - 1})$ where r, k, n $ϵ$ N is equal to

If $p, q, r$ are in A.P., then the value of determinant $∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} + 2^{n + 1} + 2 p & b^{2} + 2^{n + 2} + 3 q & c^{2} + p 2^{n} + p & 2^{n + 1} + q & 2 q a^{2} + 2^{n} + p & b^{2} + 2^{n + 1} + 2 q & c^{2} - r \end{matrix} ∣ ∣ ∣ ∣ ∣$ is

\lim_{x \to 3} \frac{\sum_{r = 1}^{n} x^{r} - \sum_{r = 1}^{n} 3^{r}}{x - 3}

is real to

(a)

\frac{(2 n - 1) \times 3^{n}}{4}

(b)

\frac{(2 n - 1) \times 3^{n} + 1}{4}

(c)

(2 n - 1) 3^{n} + 1

(d)

\frac{(2 n - 1) \times 3^{n} - 1}{4}

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The value of ∑nr=1nPrr!

The correct option is B 2n−1LHS =∑nr=1nPrr!⇒nCr=nPrr!∑nr=1nCr=nC1+nC2.........nCn =nC0+nC1+nC2.........nCn−nC0=2n−1

The value of $\sum_{r = 1}^{n} \frac{^{n} P_{r}}{r!}$