Let
XA,XB,XC and
XD represent number of balls present in box
A,B,C and
D respectively.
As no box can be empty so,
xA≥1,xB≥1,xC≥1 and
XD≥1⇒xA−1≥0,⇒xB−1≥0,⇒xC−1≥0 and
⇒xD−1≥0tA≥0,tB≥0,tC≥0 and
tD≥0 According to the question,
XA+XB+XC+XD=10⇒(XA−1)+(XB−1)+(XC−1)+(XD−1)=6
⇒tA+tB+tC+tD=6
Now question becomes, box A, B, C, and D can have none or one or more balls and total balls are 6
From formula we know,
n things can be distributed among r people in
n+r−1Cr−1 ways where each people can have either 0 or more things.
∴6 balls can be distributed among 4 boxes in
6+4−1C4−1=9C3 ways where each box can have either 0 or more balls.
Therefore, Statement 1 is correct. The number of ways of choosing any 3 places from 9 different places is
9C3 ways. But Statement - 2 is not the correct explanation of Statement -
1.