In how many ways can $12$ distinct pens be divided equally
1) among $3$ children.
options
a) 12!/$(4!)^3$ b) 12!/$(4!)^3$ . 3! c) 12!/$(5!)^3$ d) 11!/$(4!)^3$
2) into $3$ parcels.
options
a) 12!/$(4!)^3$ b) 12!/$(4!)^3$ . 3! c) 12!/$(5!)^3$ d) 11!/$(4!)^3$
My approach:
I know here how they to divide them among 3 children by keeping $4$ in each group
HOW i think:
First Select $4$ from $12$,then from remaining again select $4$ and then again do the same thing.
$12C4$ . $8C4$ . $4C4$
=$12!$/$(4!)^3$
$\endgroup$ 3Can anyone give me the HINT on How to solve the second part and also I am confused what difference it has with respect to Ist part?
1 Answer
$\begingroup$Hint for second part:
If you want to divide the pens equally among the 3 children, you can
$\;\;\;$1) divide the pens into 3 equal parcels, and then
$\;\;\;$2) distribute the parcels to the children, which can be done in $3!$ ways.
Therefore the answer to part a) can be obtained by multiplying the answer to part b) by $3!$,
as juantheron's previous answer indicated.
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