Question

The number of ways in which $5$ different toys can be distributed to $3$ children if each child can get any number of toys, is also equal to

• A. Number of subsets of $A=\{1,2,3,4,5,6,7,8,9,10\}$ which contain $2$ elements of $A$ such that their sum not equal to $11$
• B. Number of non-negative integral solutions of the equation $xyz=2310$
• C. Number of $6-$digit numbers less than $200000$ formed by using only the digits $1,2$ and $3$
• D. Number of all possible selections of one or more questions from $5$ given questions, with an alternative question corresponding to each question

Answer 1

Answer: (A), (B), (C)

Number of $6-$digit numbers less than $200000$ formed by using only the digits $1,2$ and $3$

Number of ways in which $5$ different toys can be distributed to $3$ children if each child can get any number of toys $=3^5$

$\to$ There are $5$ pairs of numbers having sum equal to $11$ i.e.,
$(1,10);(2,9);(3,8);(4,7);(5,6)$
For no two elements of the subset to have sum equal to $11:$
For each pair, we can select one number or not select both the numbers. ($3$ ways)
So, required number of subsets $=3^5$

$\to$ $xyz=2310=2\cdot 3\cdot 5\cdot 7\cdot 11$
$2$ can be distributed to $x,y,z$ in $3$ ways.
$3$ can be distributed to $x,y,z$ in $3$ ways.
$5$ can be distributed to $x,y,z$ in $3$ ways
$7$ can be distributed to $x,y,z$ in $3$ ways.
$11$ can be distributed to $x,y,z$ in $3$ ways.
Hence, total number of ways is $3^5$

$\to$ $1,-,-,-,-,-$
$1$st place can be filled only in one way, i.e., $1$
Remaining $5$ places can be filled in $3^5$ ways.
Hence, total number of ways $=1\times 3^5=3^5$

$\to$ For every question, we have three choices $:$ to do that question or its alternative or not to do the question.
So, number of all possible ways is $(3{)}^5$ as there are $5$ questions and $3$ choices for each question.
But we have to select at least one question.
So there will be $3^5-1$ ways.