Question

1. The value of sigma r=1 to n nPr/r! is
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Answer 1

SOLUTION

TO DETERMINE

The value of

$\displaystyle \sum\limits_{r=1}^{n} \: \: \frac{ {}^{n} P_r }{r \: !}$

EVALUATION

Here the given expression is

$\displaystyle \sum\limits_{r=1}^{n} \frac{ {}^{n} P_r }{r \: !}$

We now simplify it

$\displaystyle \sum\limits_{r=1}^{n} \frac{ {}^{n} P_r }{r \: !}$

$= \displaystyle \sum\limits_{r=1}^{n} \frac{n\: ! }{(n - r) ! \: r \: !}$

$= \displaystyle \sum\limits_{r=1}^{n} \: \: {}^{n}C_r$

Now we are aware of the Binomial Expansion that

${(1 + x)}^{n} = 1 + {}^{n}C_1 \: x + {}^{n}C_2 {x}^{2} + .... + {}^{n}C_n \: {x}^{n}$

Putting x = 1 we get

${(1 + 1)}^{n} = 1 + {}^{n}C_1 + {}^{n}C_2 + .... + {}^{n}C_n \:$

$\implies 1 + {}^{n}C_1 + {}^{n}C_2 + .... + {}^{n}C_n = {2}^{n} \:$

$\implies {}^{n}C_1 + {}^{n}C_2 + .... + {}^{n}C_n = {2}^{n} - 1 \:$

$\implies \: \: \displaystyle \sum\limits_{r=1}^{n} \: \: {}^{n}C_r = {2}^{n} - 1$

Hence

$\displaystyle \sum\limits_{i=1}^{n} \: \: \frac{ {}^{n} P_r }{r \: !}$

$= \displaystyle \sum\limits_{r=1}^{n} \: \: {}^{n}C_r$

$= {2}^{n} - 1$

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Answer 2

Answer:

2^n-1

Step-by-step explanation:

answer is in the attachment below