In combination and permutation what should I do if he says repetition "allowed" && order doesn't matter or anything with the same meaning. I am thinking of multiply it by a number to remove the repetition possibilities but I can't seem to get it right. example we have 7 men and 4 women in how many ways can they acquire three jobs knowing that one person can take more than one task (order doesn't matter)
edited the mistake my bad
There exist well known formulas for repetition and non-repetition for combination and permutation. Then you need to analyze the question and depending in the presented case (ordering, repetitions, length) choose one of these formulas:
Combination without repetition of $n$ elements taken $k$ by $k$. Order doesn't matter, so $abc$ is the same as $cba$: $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$
Combination with repetition of $n$ elements taken $k$ by $k$ where elements can be repeated once or more. Order doesn't matter ($abc=cba$) and you can also have $aab$, $aaa$ etc: $$\binom{n+k-1}{k} = \frac{(n+k-1)!}{k!((n+k-1)-k)!} =\frac{(n+k-1)!}{k!(n-1)!}$$
Variation without repetition of $n$ elements taken $k$ by $k$. This is a permuted combination, so ordering does matter ($abc\neq cba$ both are different):
$$V_{n,k} = \frac{n!}{(n-k)!}$$
Variation with repetition of $n$ elements taken $k$ by $k$. This is the same as a variation but with the condition that elements can be repeated:
$$V_{R_{n,k}} = n^k$$
Permutation without repetition of $n$ elements. Here we have $n=k$ so the length of the arrangements is equal to the amount of elements: $$P_n=n!$$
Permutation with repetition of $n$ elements with $a,b,c\cdots k$ elements repeated: $$P_{n_{,a,b\cdots k}} = \frac{n!}{a!b!c!\cdots k!}$$