How many different 3 digit numbers can be formed by using the digits 1 , 1 , 3 , 4 , 5
\[5P3 \]That'd be when there aren't 2 ones.
its not that easy :)
There are four choices for the first since two digits are the same, right?
but there are two one's in which is repeating
\[\LARGE{\frac{^5P_3}{2 ! }}\]
that is not right
wt is the answer?
I m also getting confused.
33
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?
In first solution; I was getting 30 as answer & In second one; I m getting 36 but not 33:\
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.
where p,q,r,s are the number of items in each group.
ya but how to get " in how many ways 3 digit numbers would be formed?"
Ohh, that's true! Silly me! I'll have to work harder.
me too hehe;)
take two cases
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.
1. when all d digits are different...1,3,4,5 total no of 3digits=24
case 2. when there are two 1's in three digit no
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
so total no of three digits formed by 1,1,3,4,5= 24+9=33
for case 1, how is the total number of digits 24?
4 p 3 =24
you have 4 different digits and have to make 3 digit nos from them..
i got it now. thanks
:)
nice:)
Yep, thanks.
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