Let A denote the event that a $$6-digit$$ integer formed by $$0$$, $$1$$, $$2$$, $$3$$, $$4$$, $$5$$, $$6$$ without repetitions, be divisible by $$3$$. Then probability of event A is equal to:

Question

Let A denote the event that a $$6-digit$$ integer formed by $$0$$, $$1$$, $$2$$, $$3$$, $$4$$, $$5$$, $$6$$ without repetitions, be divisible by $$3$$.  Then probability of event A is equal to:

Infinity Learn · Accepted Answer

Total number of $$6$$ digited numbers from $$0$$, $$1$$, $$2$$, $$3$$, $$4$$, $$5$$, $$6=6$$×$$6$$×$$5$$×$$4$$×$$3$$×$$2$$ ∴$$n(S)=6$$×$$6$$!;    We have $$0+1+2+3+4+5+6=21Therefore$$ $$6$$ digited number divisible by $$3$$ can be formed by using the digits  $$1$$,$$2$$,$$3$$,$$4$$,$$5$$,$$6$$ in $$6$$! ways.  By using  the digits $$0$$,$$1$$,$$2$$,$$4$$,$$5$$,$$6$$ in $$5$$×$$5$$!  ways by using the digits $$0$$,$$1$$,$$2$$,$$3$$,$$4$$,$$5$$ in $$5$$×$$5$$! ways  Hence, $$P(E)=6$$!$$+2$$×$$5$$×$$5$$!$$6$$×$$6$$!$$=6+1036=49$$

$L e t A d e n o t e t h e e v e n t t h a t a 6 - d i g i t i n t e g e r f o r m e d b y 0, 1, 2, 3, 4, 5, 6 w i t h o u t r e p e t i t i o n s, b e d i v i s i b l e b y 3 . T h e n p r o b a b i l i t y o f e v e n t A i s e q u a l t o :$

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Let A denote the event that a 6-digit integer formed by 0, 1, 2, 3, 4, 5, 6 without repetitions, be divisible by 3. Then probability of event A is equal to:

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$L e t A d e n o t e t h e e v e n t t h a t a 6 - d i g i t i n t e g e r f o r m e d b y 0, 1, 2, 3, 4, 5, 6 w i t h o u t r e p e t i t i o n s, b e d i v i s i b l e b y 3 . T h e n p r o b a b i l i t y o f e v e n t A i s e q u a l t o :$