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johri_anshuman

How many 6 digit numbers are there such that they fulfill the following 2 conditions
a) The digits of each number are from {1,2,3,4,5}
b) Any digit that appears in the number appears at least twice (eg. 225252 is admissible while 222133 is not)

krishna.gopal

This thin is possible in two ways
1) There are 3 digits repeated twice
2) There are 1 digit which comes 4 times and another which comes twice.
For 1)
Number of ways of choosing 3 digits 5C3
Number of ways of arranging them = 6!/(2!)^3
For 2)
Number of ways of choosing 2 digits 5C2
Two ways to decide which one will come 4 times and which one will come twice
Number of ways of arranging them = 6!/(2!*4!)

so answer is 5C3*6!/(2!)^3+5C2*2*6!/(4!*2!)

iitkgp_bipin

Third case is also possible in which a digit comes 6 times.

For this case, the no. ways are 5.
(five possible numbers are 111111,222222,333333,444444,555555 ; hence 5 possibilities).

Hence 5 should be added to Krishna Sir's answer ot get the total number of 6 digit numbers.

sandesh1

i also appear that exam.

& i calculated answer is 960.

my friend's opinion is answer is above 500.

so, 960 may be correct??????

do, rate me, if correct!!!!!!!!

sandymaurya

Since every no. (from 1 to 5) can be repeat so there are 10 no which can be used.
i.e. 1,1,2,2,3,3,4,4,5,5

but we need only a no. containg 6 digit out of 10 digits in which three are repeated.
Or we can say the total no. of combinations which can be formed are from five digits and we have to choose only three digits, because when we choose a digit say 1 than by condition we should have to choose another 1.
In sort you can say we have to choose in pairs.
Thus total no. of combinations possible are ⁵C_{3 = 10.

Now for each combination we have maximum arrangements with repetitions are 6!.} So we have a formula for repeated permutations.

(no. of possible outcomes with repetitions) / p_{1! *} p_{2! *} ... _* p_n!

where p_1, p_2, and p_n are repeatitions

thus for each combination

P₁ = 6!/(2!_*2!_*2!)=630

and for 10 combinations

P = 10 x P₁ = 10 x 6! / ( 2! x 2! x 2! ) = 100

IF U FIND THIS USEFUL FOR U AND ALSO IF IT IS CORRECT THAN RATE IT

johri_anshuman

but one more case is possible

one digit comes 3 times and another digit also comes three times

so ways of selecting there 2 digits=5C2=10
number of ways of permuting these digits=6!/3!3!=720/36=20

so total such numbers=200

So adding all cases

total numbers = 5C3*6!/(2!)^3+5C2*2*6!/(4!*2!) + 5 +200
=900+300+5+200=1405

nadeemoidu

@johri_anshuman

How did you get 5C3*6!/(2!)^3 as 3600 ????

It is 900

So I think the answer is 900 + 300 + 200 + 5 = 1405

or atleast that's what I wrote for the olympiad :)

goutam.chalasani

nadeemoidu is right
even i got the same answer

rajatsen91

look at the problem in this way
there can be three pairs
case 1: all different
no of numbers in this case = 5C3*6!/(2!)^3 = 900
case 2 : two pairs same other different
total = 5C1*4C1*6!(4!*2!) = 300
case 3 : all alike
no of possibilities = 5
total = 1205

krzme

even I got 1405!

What about:
Prove that
(a) 5< 5^0.5 + 5^0.33 + 5^0.25;
(b) 8 > 8^0.5 + 8^0.33 + 8^0.25;

krzme

also
Q1:
Let ABC be an acute angled triangle; AD be the bisector of angle BAC with D on BC; and BE be the altitude from B on AC.
Show that angle CED>45 degree.

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