15) How many different numbers of six digits without repetition can be formed from the digits 0,1,,3,5,7,9?

1. How many of them will have 0 in the unit place?
2. How many of them are divisible by 5?
3. How many of them are not divisible by 5?

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} 0,1,3,5,7,9, \\ \mathrm{Total} 6 \mathrm{digits} \mathrm{numbers}=\boxed{5} \boxed{5} \boxed{4} \boxed{3} \boxed{2} \boxed{1} \\ \mathrm{We} \mathrm{will} \mathrm{have} 5 \mathrm{options} \mathrm{to} \mathrm{fill} \mathrm{first} \mathrm{number} \mathrm{as} 0 \mathrm{cannot} \mathrm{be} \mathrm{filled} \mathrm{in} \mathrm{the} \mathrm{first} \mathrm{place}. \\ \Rightarrow \mathrm{Total} \mathrm{numbers}=5\times 5\times 4\times 3\times 2\times 1=\mathbf{600}\mathbf{ }\left(\mathbf{Answer}\right) \\ \mathrm{To} \mathrm{get} \mathrm{numbers} \mathrm{with} 0 \mathrm{as} \mathrm{unit} \mathrm{digits}, \mathrm{fix} 0 \mathrm{as} \mathrm{unit} \mathrm{digit}, \mathrm{and} \mathrm{fill} \mathrm{remaining} \mathrm{numbers}. \\ \boxed{5} \boxed{4} \boxed{3} \boxed{2} \boxed{1} \boxed{1} , \mathrm{there} \mathrm{is} \mathrm{only} \mathrm{one} \mathrm{way} \mathrm{to} \mathrm{fill} \mathrm{last} \mathrm{digit} \mathrm{i}.\mathrm{e}. \mathrm{to} \mathrm{fill} 0. \\ \Rightarrow \mathrm{Total} \mathrm{numbers} \mathrm{with} 0 \mathrm{as} \mathrm{unit} \mathrm{digit}=5\times 4\times 3\times 2\times 1\times 1=\mathbf{120}\mathbf{ }\left(\mathbf{Answer}\right) \\ \mathrm{Now} \mathrm{total} \mathrm{numbers} \mathrm{with} 5 \mathrm{as} \mathrm{unit} \mathrm{digit} \mathrm{will} \mathrm{be} \mathrm{given} \mathrm{by} \mathrm{fixing} \mathrm{unit} \mathrm{digit} \mathrm{as} 5. \\ \boxed{4} \boxed{4} \boxed{3} \boxed{2} \boxed{1} \boxed{1} , \mathrm{first} \mathrm{place} \mathrm{cannot} \mathrm{be} \mathrm{filled} \mathrm{by} 0 \mathrm{and} 5. \\ \Rightarrow \mathrm{Total} \mathrm{numbers} \mathrm{with} 5 \mathrm{as} \mathrm{unit} \mathrm{digit}=4\times 4\times 3\times 2\times 1\times 1=96 \\ \mathrm{Numbers} \mathrm{which} \mathrm{are} \mathrm{divisible} \mathrm{by} 5 \mathrm{will} \mathrm{have} 0 \mathrm{or} 5 \mathrm{at} \mathrm{unit} \mathrm{place}. \\ \Rightarrow \mathrm{Total} \mathrm{numbers} \mathrm{divisible} \mathrm{by} 5=120+96=\mathbf{216}\mathbf{ }\left(\mathbf{Answer}\right) \\ \mathrm{Total} \mathrm{numbers} \mathrm{not} \mathrm{divisible} \mathrm{by} 5=\mathrm{Total} \mathrm{numbers}-\mathrm{Total} \mathrm{numbers} \mathrm{divisible} \mathrm{by} 5=600-216 \\ \Rightarrow \mathrm{Total} \mathrm{numbers} \mathrm{not} \mathrm{divisible} \mathrm{by} 5=\mathbf{384}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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