IIT JEE , AIEEE Forum - Mathematics , Algebra - Good question of Permutations & combination.

tarinbansal

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.

nik_gup

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

rahuljyotinath

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

learner

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

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

tarinbansal

Yes Rahul, this question is from RMO-2007.

tarinbansal

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

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

Mystic00

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

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

btetarbe

For RMO 2007 solutions open the following website:

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

bvsatyaram

Yes.. Priyesh is right.

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