Permutations & Combinations
Distribution and Stars and Bars
GRB_1000_MCQ
Grade Class 12
Question:
50 identical marbles are to be distributed among four boys, $A_1$, $A_2$, $A_3$ and $A_4$. The number of marbles receiving by them in the distribution are as follows:
$$A_1: 1, 3, 5, 7, \ldots$$
$$A_2: 4, 6, 8, 10, \ldots$$
$$A_3: 5, 7, 9, 11, \ldots$$
$$A_4: 2, 4, 6, 8, \ldots$$
Identify which of the following statement(s) is(are) <b>correct</b>?
The total number of ways of distribution is $^{22}C_3$
The total number of ways of distribution is $^{20}C_3$
If $A_4$ is receiving not more than 14 marbles, then number of ways of distribution is 960.
If $A_4$ is receiving not more than 14 marbles, then number of ways of distribution is 1085.
Step-by-Step Solution
Step 1: Express each boy's marble count in terms of a variable. Let $A_1$ receive $(2a-1)$ marbles where $a \geq 1$, $A_2$ receive $(2b+2)$ marbles where $b \geq 1$, $A_3$ receive $(2c+3)$ marbles where $c \geq 1$, and $A_4$ receive $2d$ marbles where $d \geq 1$.
Step 2: Set up the equation for total marbles.
$$(2a-1) + (2b+2) + (2c+3) + 2d = 50$$
$$2a + 2b + 2c + 2d + 4 = 50$$
$$2(a+b+c+d) = 46$$
$$a + b + c + d = 23$$
Step 3: Count the number of positive integer solutions using stars and bars.
The number of solutions in positive integers is $^{23-1}C_{4-1} = ^{22}C_3$. So option (1) is correct.
Step 4: For the condition that $A_4$ receives not more than 14 marbles, i.e., $2d \leq 14 \Rightarrow d \leq 7$.
Total solutions without restriction: $^{22}C_3 = 1540$.
Solutions where $d \geq 8$: Let $d' = d - 7$, so $a + b + c + d' = 16$ with $a,b,c,d' \geq 1$, giving $^{15}C_3 = 455$ solutions.
So valid solutions $= 1540 - 455 = 1085$. So option (4) is correct.
Correct Answer: 1, 4