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Question Number 883 by 112358 last updated on 12/Apr/15

$$Howmanyevennumbers$$$$between3000and7000canbe$$$$formedusingthedigits$$$$2,3,4,5,6and8ifrepetition$$$$ofdigitsisnotallowed?$$

Answered by prakash jain last updated on 12/Apr/15

$$\mathrm{First}\mathrm{digits}3,5=2$$$$\mathrm{Last}\mathrm{digit}(2,4,6,8)=4$$$$\mathrm{Middle}\mathrm{two}\mathrm{digits}{}^4{\mathrm{P}}_2=\frac{4!}{2!}=12$$$$\mathrm{Total}\mathrm{with}\mathrm{first}\mathrm{digits}3,5=2\times 4\times 12=96$$$$\mathrm{First}\mathrm{digit}4,6=2$$$$\mathrm{Last}\mathrm{digits}(2,6,8)\mathrm{or}(4,6,8)=3$$$$\mathrm{Middle}\mathrm{two}\mathrm{digits}{}^4{\mathrm{P}}_2=\frac{4!}{2!}=12$$$$\mathrm{Total}\mathrm{with}\mathrm{first}\mathrm{digits}4,6=2\times 3\times 12=72$$$$\mathrm{Total}=72+96=168$$

Commented by 112358 last updated on 12/Apr/15

$$Thanks!$$

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