71Of Course, But won;t it be tedious?
Say x=y=1. So z ≤ 26 , Hence 26 Solutions
Say x=1,y=2, So z ≤ 25 Hence 25 Solutions
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1) Find the total no.of +ve integral solutions of the inequality 3x+y+z<= 30..
Can a student not knowing multinomial theorem solve this Problem??
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11 Answers
71
21We can shorten it up.
Let p be a non-negative integer such that 3x+y+z+p=30
Now apply brute force.
Case No. of solutions
x=1 29C2
x=2 26C2
x=3 23C2
. .
. .
. .
x=10 2C2
Now, Total no. of cases=1495
1057@arnab but the question says no of +ve integral solutions..
so fixing x:
Case No. of solutions
x=1 26
x=2 23
x=3 20
x=4 17
x=5 14
x=6 11
x=7 8
x=8 5
x=9 2
x=10 0
Note:in all the cases i have used the result that no of +ve integral solutions of x1+ x2....upto r terms =n is (n-1)C(r-1)
the answer is 126...
1057oh!i forgot......wat i posted is only the solution for the equation being exactly 30
dint see the ≤ sign....
neways for 29 we have 25+22+19.....+1
for 28 we have 24+21+18....+3
for 27 we have 23+20+.....2
for 26 we have 22+19......+1
and so on we do till 3..
add them all up(its not that difficult as they are in AP)....
62There is a still simpler solution
3x+y+z<= 30..
is same as 3x+y+z+w=30
where w is a non negative integer....
x=x'+1
y=y'+1
z=z'+1
this makes the solutions non negative instead of +ve integer
y'+z'+w=25-3x'
now solve case by case