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The HR nominates 12 computer engineers, 8 civil engineers and 20 electrical engineers for the transfer to the site office. But 5 among them cannot go as they are engaged in the on-going arbitration and another 3 already opted for transfer. In how many ways can the transfer list of 10 head counts be made now?
Option: 1
$\frac{35!}{18!17!}$

Option: 2
$\frac{32!}{7!25!}$

Option: 3
$\frac{35!}{19!17!}$

Option: 4
Cannot be determined

Answers (1)

best_answer

Note the following:

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

Since 5 engineers will be excluded and 3 already opted for transfer, the following is evident.

The number from which the restricted combination is to be made is $\mathrm{=n-k-h=\left ( 12+8+20 \right )-3-5=32}$ .
The number with which the restricted combination is to be made is $\mathrm{=r-k=10-3=7}$

Therefore, the required restricted combination is

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

$\mathrm{=^{32}C_{7}}$

$=\frac{32!}{7!25!}$

Posted by

Ajit Kumar Dubey

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