Answer:
The answer is "0.30%"
Step-by-step explanation:
To assign employees to work, we have to select 6 out of 20 to do work, instead select 4 out of another 14 jobs, 2 to do, and 5 out of another 10 works 3. The other five were going to go do work four. They may decide how often ways of making each of these options. Their service would be how many forms to allocate those four positions to staff. The random sample is from the height.
There are equally estimated that numbers of arrangements with both the four in Job 1.They then allocate all four to work 1. In this situation We must then select 2 out from the 16 existing spaces in job 1, 4 out from the 14 remaining spaces in job 2, 5 out of the 10 existing spaces in job 3 as well as the ones remaining in job 4.
The possibility of any four being at 1 is that they've been all divided by the overall amount of circumstances in 1 task.
[tex]\to \frac{[ \frac{16!}{(2!4!5!5!)}]}{[\frac{20!}{(6!4!5!5!)}]}\\\\\to [ \frac{16!}{(2!4!5!5!)}] \times [\frac{(6!4!5!5!)}{20!}]\\\\\to [ \frac{16!}{(2!)}] \times [\frac{(6!)}{20!}]\\\\\to [ \frac{16!}{(2!)}] \times [\frac{(6\times 5 \times 4 \times 3 \times 2!)}{20!}]\\\\\to 16! \times [\frac{(360)}{20 \times 19 \times 18 \times 17 \times 16!}]\\\\\to \frac{(360)}{20 \times 19 \times 18 \times 17 }\\\\\to \frac{(360)}{20 \times 19 \times 18 \times 17 }\\\\\to \frac{360}{116,280}[/tex]
[tex]\to 0.00309597523 \approx 0.30 \%[/tex]