Question

In how many ways can you partition $6$ into ordered summands? (For example, $3$ can be partitioned in $3$ ways as : $1+2,2+1,1+1+1$)

• A. $27$
• B. $29$
• C. $31$
• D. $33$

Answer 1

Answer: (C)

$31$

$1+1+1+1+1+1$ $1$ ways

$2+1+1+1+1$ $5$ ways
$3+1+1+1$ $4$ ways
$4+1+1$ $3$ ways
$5+1$ $2$ ways
$2+3+1$ $6$ ways
$2+4$ $2$ ways
$3+3$ $1$ ways
$2+2+2$ $1$ ways
$2+2+1+1$ $6$ ways

Total number of ways of partitioning $6=1+5+4+3+2+6+2+1+1+6=31$.

Hence, option C is correct.