Statement - 1: The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is emply is 9C3.
Statement - 2: The number of ways of choosing any 3 places from 9 different places is 9C3.
Solutions · 1
Official solution
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 \[\ge\] 1, XB \[\ge\] 1, XC \[\ge\] 1 and XD \[\ge\] 1
\[\Rightarrow\] XA - 1 \[\ge\] 0, \[\Rightarrow\] XB - 1 \[\ge\] 0, \[\Rightarrow\] XC - 1 \[\ge\] 0 and \[\Rightarrow\] XD - 1 \[\ge\] 0
tA \[\ge\] 0, tB \[\ge\] 0, tC \[\ge\] 0 and tD \[\ge\] 0
According to the question,
XA + XB + XC + XD = 10
\[\Rightarrow\] (XA - 1) + (XB - 1) + (XC - 1) + (XD - 1) = 6
\[\Rightarrow\] 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.
\[\therefore\] 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.