Question Sample Titled 'Addition rule of counting'

 (a) Suppose a task can be performed by ${k}$ methods, in which method ${1}$ can be performed in ${n}_{{1}}$ ways, method ${2}$ can be performed in ${n}_{{2}}$ ways, … method ${k}$ can be performed in ${n}_{{k}}$ ways, then the number of ways to perform the task is ${n}_{{1}}+{n}_{{2}}+$ … ${n}_{{k}}$ .  (b) Suppose a task can be performed by either method ${1}$ in ${n}_{{1}}$ ways or method ${2}$ in ${n}_{{2}}$ ways, and among these ${n}_{{1}}+{n}_{{2}}$ ways, there are ${m}$ ways in common, then the number of ways to perform the task is ${n}_{{1}}+{n}_{{2}}-{m}$ .

 Example On a bookshelf, there are ${6}$ Chinese books and ${4}$ English books. If Peter wants to take a book from the bookshelf, how many choices are there? Solution By the addition rule of counting, the number of choices $={6}+{4}$ $={10}$

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