in an exam, max marks for 3 papers are 50 each. max marks for 4th paper is 100. find the no.of ways with which a student can score 60% marks in aggregate?
if he scores maximum marks or zero marks than 5 ways.
other wise cannot be determined. since he may score any marks in any subject.
bhai kuch aise nhi hoga isme ? :a+b+c+d = 120Total solutions 123C3Now we have to subtract the cases where each of a,b or c can have the values from 51 to 120, isme mein atak raha hu @amresh_maverick aap bhi dekho jara
bhai kuch aise nhi hoga isme ? :a+b+c+d = 120Total solutions 123C3Now we have to subtract the cases where each of a,b or c can have the values from 51 to 120, isme mein atak raha hu @amresh_maverick aap bhi dekho jara
it shud be a+b+c+d=150
and remove the case when a,b,c wil be 51 to 150 and cases when d wil be 101 to 150
in an exam, max marks for 3 papers are 50 each. max marks for 4th paper is 100. find the no.of ways with which a student can score 60% marks in aggregate?
p1,p2,p3
p4
p1 + p2 + p3 + p4 = (0.6)*(250) = 150
Normal ways = 153C3
But we need to remove cases when p1,p2,p3>50 or/and p4 >100
Case 1: p4 > 100
p4 = 101 + a (where a>=0)
a + p1 + p2 + p3 = 49 => Number of ways = 52C3
Case 2: p1 or p2 or p3 > 50
p1 = 51 + a
a + p2 + p3 + p4 = 99 => Number of ways = 102C3
But, there may be cases when both p1,p2 are more than 50
=> a + b + p3 + p4 = 48 => Number of ways = 51C3
Now, we need to use the concept of set theory
All - (When p4 >100) - (When p1 /p2/p3 > 50) + ( When any two of p1/p2/p3 are more than 50)
=> 153C3 - 52C3 - 3*(102C3) + 3C2*(51C3) I think..
frm a set of 3 capital and 5 small consonants and 4 small vowels, how many wrds can be formed each starting with a capital consonant and containing 3 small consonants and two small vowels?
frm a set of 3 capital and 5 small consonants and 4 small vowels, how many wrds can be formed each starting with a capital consonant and containing 3 small consonants and two small vowels?