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In how many ways can 10 packets of 1100 gemstones be made from 1600 gemstones such that 250 red gemstones, 160 blue gemstones and 180 gemstones must always be included in the 10 packets?
Option: 1
$\frac{1100!}{510!500!}$

Option: 2
$\frac{1110!}{510!510!}$

Option: 3
$\frac{1010!}{510!590!}$

Option: 4
$\frac{1010!}{510!500!}$

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 $\mathrm{=^{y}C_{x}=\frac{y!}{x!(y-x)!}}$
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 $\mathrm{=^{n-k}C_{r-k}}$

In the 10 packets of 1100 gemstones, the number of gemstones that must be present is

$=250+160+180$

$=590$

Since 590 gemstones must always be included in the packets, the following is evident.

The number from which the restricted combination is to be made is $\mathrm{=n-k=1600-590=1010}$ .
The number with which the restricted combination is to be made is $\mathrm{=r-k=1100-590=510}$

Therefore, the required restricted combination is

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

$\mathrm{=^{1010}C_{510}}$

$=\frac{1010!}{510!500!}$

Posted by

Irshad Anwar

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