$a.12!\phantom{\rule{0ex}{0ex}}b.\frac{8!}{12!}\phantom{\rule{0ex}{0ex}}c.\frac{12!}{(12-8)!}\phantom{\rule{0ex}{0ex}}d.\frac{12!}{8!}$

Buszmenan
2022-09-06
Answered

What is the permutation formula for P(12,8)?

$a.12!\phantom{\rule{0ex}{0ex}}b.\frac{8!}{12!}\phantom{\rule{0ex}{0ex}}c.\frac{12!}{(12-8)!}\phantom{\rule{0ex}{0ex}}d.\frac{12!}{8!}$

$a.12!\phantom{\rule{0ex}{0ex}}b.\frac{8!}{12!}\phantom{\rule{0ex}{0ex}}c.\frac{12!}{(12-8)!}\phantom{\rule{0ex}{0ex}}d.\frac{12!}{8!}$

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xgirlrogueim

Answered 2022-09-07
Author has **13** answers

We know that, permutation formula is

$P(n,r)=nPr\phantom{\rule{0ex}{0ex}}nPr=\frac{n!}{(n-r)!}$

where, nPr=permutation

n=Total number of object.

r=number of selected objects

$\therefore P(12,8)=12{P}_{8}\phantom{\rule{0ex}{0ex}}\therefore P(12,8)=\frac{12!}{(12-8)!}$

option (c) $P(12,8)=\frac{12!}{(12-8)!}$ is correct

$P(n,r)=nPr\phantom{\rule{0ex}{0ex}}nPr=\frac{n!}{(n-r)!}$

where, nPr=permutation

n=Total number of object.

r=number of selected objects

$\therefore P(12,8)=12{P}_{8}\phantom{\rule{0ex}{0ex}}\therefore P(12,8)=\frac{12!}{(12-8)!}$

option (c) $P(12,8)=\frac{12!}{(12-8)!}$ is correct

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