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Mathematics 17 Online
OpenStudy (anonymous):

How many different 3 digit numbers can be formed by using the digits 1 , 1 , 3 , 4 , 5

Parth (parthkohli):

\[5P3 \]That'd be when there aren't 2 ones.

OpenStudy (anonymous):

its not that easy :)

Parth (parthkohli):

There are four choices for the first since two digits are the same, right?

OpenStudy (anonymous):

but there are two one's in which is repeating

OpenStudy (maheshmeghwal9):

\[\LARGE{\frac{^5P_3}{2 ! }}\]

OpenStudy (anonymous):

that is not right

OpenStudy (maheshmeghwal9):

wt is the answer?

OpenStudy (maheshmeghwal9):

I m also getting confused.

OpenStudy (anonymous):

33

OpenStudy (maheshmeghwal9):

I m only getting this much By using above 5 digits ; we can form 5 digit numbers in \(\LARGE{\frac{5!}{2!}=60ways}\) Then in how many ways 3 digit numbers would be formed?

OpenStudy (maheshmeghwal9):

In first solution; I was getting 30 as answer & In second one; I m getting 36 but not 33:\

OpenStudy (mathmate):

I vaguely remember that it is \( \frac{n!}{(p!q!r!s!} \) ways where p+q+r+s=n. Here we have \( \frac{5!}{(2!1!1!1!} = \frac{5!}{2!}\) = 60 ways as you said.

OpenStudy (mathmate):

where p,q,r,s are the number of items in each group.

OpenStudy (maheshmeghwal9):

ya but how to get " in how many ways 3 digit numbers would be formed?"

OpenStudy (mathmate):

Ohh, that's true! Silly me! I'll have to work harder.

OpenStudy (maheshmeghwal9):

me too hehe;)

OpenStudy (anonymous):

take two cases

OpenStudy (mathmate):

Let's start with 5P3=120/2=60. Out of these, we will find the patterns: X11,1X1,11X where X could be 3,4,or 5, that's overcounting by 6. So there are 54 different numbers. I don't like the way I do it, so will look for a different method to check.

OpenStudy (anonymous):

1. when all d digits are different...1,3,4,5 total no of 3digits=24

OpenStudy (anonymous):

case 2. when there are two 1's in three digit no

OpenStudy (anonymous):

s0 1,1,.. 3rd digit can be either 3 or 4 or 5 so 3 ways and these three digits can be arranged by 3!/2! ways so total no of 3 digits= 3*3=9

OpenStudy (anonymous):

so total no of three digits formed by 1,1,3,4,5= 24+9=33

OpenStudy (anonymous):

for case 1, how is the total number of digits 24?

OpenStudy (anonymous):

4 p 3 =24

OpenStudy (anonymous):

you have 4 different digits and have to make 3 digit nos from them..

OpenStudy (anonymous):

i got it now. thanks

OpenStudy (anonymous):

:)

OpenStudy (maheshmeghwal9):

nice:)

OpenStudy (mathmate):

Yep, thanks.

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