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In how many ways can a shopkeeper remove the 100 lowest-price books from the shelf of 150 books among which 50 are damaged and 30 are specimen copies?
Option: 1
$\frac{70!}{20!50!}$

Option: 2
$\frac{100!}{20!50!}$

Option: 3
$\frac{150!}{20!50!}$

Option: 4
Cannot be determined

Answers (1)

best_answer

Note the following:

The formula for the combination for the selection of the $\mathrm{x}$ items from the $\mathrm{y}$ different items is $\mathrm{=^{y}C_{x}=\frac{y!}{x!\left ( y-x \right )!}}$
The restricted combination for the selection of the $\mathrm{r}$ items from the $\mathrm{n}$ different items with $\mathrm{k}$ particular things always included is $\mathrm{=^{n-k}C_{r-k}}$

As per the available data, the number of the books which must be in the rejection list is

$=50+30$

$=80$

Since 80 books are confirmed to be removed from the shelf, the following is evident.

The number from which the restricted combination is to be made is $\mathrm{=n-k=150-80=70}$ .
The number with which the restricted combination is to be made is $\mathrm{=r-k=100-80=20}$

Therefore, the required restricted combination is

$\mathrm{=^{n-k}C_{r-k}}$

$\mathrm{=^{70}C_{20}}$

$=\frac{70!}{20!50!}$

Posted by

Ritika Harsh

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