QJEEM11Ma064d: JEE Main 2011 Mathematics - Class 11 JEEM11Ma064d Solution: Statement - 1: The number of ways of distributing 10 identical…

Let XA, X_B, X_C and X_D represent number of balls present in box A, B, C and D respectively.

As no box can be empty so,

X_A $\ge$ 1, X_B $\ge$ 1, X_C $\ge$ 1 and X_D $\ge$ 1

$\Rightarrow$ X_A $-$ 1 $\ge$ 0, $\Rightarrow$ X_B $-$ 1 $\ge$ 0, $\Rightarrow$ X_C $-$ 1 $\ge$ 0 and $\Rightarrow$ X_D $-$ 1 $\ge$ 0

t_A $\ge$ 0, t_B $\ge$ 0, t_C $\ge$ 0 and t_D $\ge$ 0

According to the question,

X_A + X_B + X_C + X_D = 10

$\Rightarrow$ (X_A $-$ 1) + (X_B $-$ 1) + (X_C $-$ 1) + (X_D $-$ 1) = 6

$\Rightarrow$ t_A + t_B + t_C + t_D = 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 ${}^{n + r - 1}{C_{r - 1}}$ ways where each people can have either 0 or more things.

$\therefore$ 6 balls can be distributed among 4 boxes in ${}^{6 + 4 - 1}{C_{4 - 1}} = {}^9{C_3}$ 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 ${}^9{C_3}$ ways. But Statement - 2 is not the correct explanation of Statement - 1.