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Algebra

Blazing goIITian

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2 Dec 2007 00:18:39 IST

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790

Good question of Permutations & combination.

None

How many 6 digit numbers are there such that any digit that appears in the number appears at least twice.

(Example-225252 is an admissible number while 222133 is not)

Numbers are to be chosen from the set {1,2,3,4,5}

I don't know the answer. Please post ur solutions.

Rates assured.

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Nikhil Gupta

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2 Dec 2007 11:36:14 IST

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i think nadeemoidu has posted the link for solution of ur query in quadratic eqn problem

rahul nath

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2 Dec 2007 13:40:52 IST

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isn't that from rmo-2007
i appeared in it.
wait for soln

Learner .

Blazing goIITian

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2 Dec 2007 18:17:28 IST

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Each no.comes atleast twice:

Case a)Three nos. twice each

Take three numbers from the given set,lets take 1,2,3:

Now ways of arranging them :6P3/2*2*2

Since three are to be chosen for this purpose,we have5C3.

Total:6P3*5C3

Case b)Two nos thrice each:

Same way as above,6P2*5c2/3*3

Case c)One no:

no. of cases: 5

Adding everything,I get 255.

Please tell me if I've made an error !

Priyesh

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2 Dec 2007 18:29:00 IST

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since each digit should appear atleast twice

so 4 cases are possible

case 1: three digits appear two times each

so no. of ways of selcting those three digits = 5 C 3 = 10

no . of of arranging = 6!/2!*2!*2! = 90

so total nos = 10 * 90 = 900

case 2: two digits appear three times each

so no. of ways of selecting = 5 C 2 = 10

no. of ways of arranging = 6!/3!*3! = 20

so total nos. = 200

case 3: one digit appears four times & the other two times

no of ways in which the no. which appears four times can be selected is 5C1 = 5 ways

no . of ways in which the other no. can be selected from the remaining set of four nos. is 4C1 = 4 ways

arrangement can be done in 6!/4!*2!

so total nos = 5 * 4 * 6!/4!*2! = 300

case 4: all digits are same

total nos 5

so adding all cases total nos = 900 + 200 + 300 + 5 = 1405

Tarin Bansal

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2 Dec 2007 19:31:34 IST

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Yes Rahul, this question is from RMO-2007.

Tarin Bansal

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2 Dec 2007 19:33:57 IST

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Priyesh,
your solution seems to be correct, But my doubt is that total number of numbers are coming out to be 15625 and ur answer is 1405,
Can there be such a large difference??????
Because only a few cases are neglected w.r.t to total numbers in this question. Only those cases are neglected which have every number only once.
SO are there as many cases as 14220?????
I don't think so.

Priyesh

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2 Dec 2007 20:24:52 IST

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hey tarin

a lot of cases are neglected see

nos in which two nos. occur twice & two other nos.( amounts to 5400)

nos. in which one digit occurs twice & four other nos.(amounts to 1800)

so u can see that many cases more like these will add up to give 14220

Paresh Manchandia

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2 Dec 2007 20:59:12 IST

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I dont think tarin has missed any case in the solution
even i got the ans 1405

@priyesh

nos in which two nos. occur twice & two other nos.( amounts to 5400)

there are no other 2 nos. because they have to be the same (read ques)

nos. in which one digit occurs twice & four other nos.(amounts to 1800)

again there are no 4 other nos. each no has to appear 2 or more times

Priyesh

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2 Dec 2007 21:17:50 IST

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hey mystic yaar i was replying to the doubt of tarin(tarin asked me that have i missed any case or not) so i wrote the cases not favourable to the condition(and not the ones favourable)

look at my solution above i got the ans as 1405

Bhaskar Tetarbe

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2 Dec 2007 22:36:13 IST

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For RMO 2007 solutions open the following website:

http://www.isid.ac.in/~rbb/crmosol_07.pdf

B.V. Satyaram

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Joined: 10 Dec 2007

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11 Dec 2007 00:59:49 IST

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Yes.. Priyesh is right.

sakshi jain

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26 Nov 2011 17:48:38 IST

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Good question of Permutations & combination.

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