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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.

$12C4$ . $8C4$ . $4C4$

=$12!$/$(4!)^3$

Can 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?

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    $\begingroup$ I don't understand what the difference would be between (1) and (2). In each case you are dividing them into 3 groups. $\endgroup$ – Paul Oct 1 '15 at 16:55
  • $\begingroup$ @Paul Same thought I have too. $\endgroup$ – Jack Oct 1 '15 at 16:56
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    $\begingroup$ My guess is that when you divide them amongst three children, the children are to be distinguished from one another, and when you divide them into three parcels, the parcels are indistinguishable. $\endgroup$ – Brian Tung Oct 1 '15 at 17:04
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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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