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Question Number 102139 by mathocean1 last updated on 06/Jul/20

We have 4 boys and 2 girls. we  choose at random and simultaneous  2 boys and 1 girl to for a group.  1) How many possibilities do we have?

$${We}\:{have}\:\mathrm{4}\:{boys}\:{and}\:\mathrm{2}\:{girls}.\:{we} \\ $$$${choose}\:{at}\:{random}\:{and}\:{simultaneous} \\ $$$$\mathrm{2}\:{boys}\:{and}\:\mathrm{1}\:{girl}\:{to}\:{for}\:{a}\:{group}. \\ $$$$\left.\mathrm{1}\right)\:{How}\:{many}\:{possibilities}\:{do}\:{we}\:{have}? \\ $$

Commented by prakash jain last updated on 07/Jul/20

^4 C_2 ×^2 C_1 =24.

$$\:^{\mathrm{4}} {C}_{\mathrm{2}} ×^{\mathrm{2}} {C}_{\mathrm{1}} =\mathrm{24}. \\ $$

Answered by geoplitz last updated on 06/Jul/20

Boys: b_1 , b_2 , b_3 , b_4   Girls: g_1 , g_2   With b_1  and b_2 , we have 2 possible groups.  (b_1 , b_2 ) → 2  So, we have  (b_1 , b_2 ) → 2            (b_2 , b_3 ) → 2  (b_1 , b_3 ) → 2            (b_2 , b_4 ) → 2  (b_1 , b_4 ) → 2            (b_3 , b_4 ) → 2  So, we have 12 possibilities.  2 + 2 + 2 + 2 + 2 + 2 = 12

$$\mathrm{Boys}:\:{b}_{\mathrm{1}} ,\:{b}_{\mathrm{2}} ,\:{b}_{\mathrm{3}} ,\:{b}_{\mathrm{4}} \\ $$$$\mathrm{Girls}:\:{g}_{\mathrm{1}} ,\:{g}_{\mathrm{2}} \\ $$$$\mathrm{With}\:{b}_{\mathrm{1}} \:\mathrm{and}\:{b}_{\mathrm{2}} ,\:\mathrm{we}\:\mathrm{have}\:\mathrm{2}\:\mathrm{possible}\:\mathrm{groups}. \\ $$$$\left({b}_{\mathrm{1}} ,\:{b}_{\mathrm{2}} \right)\:\rightarrow\:\mathrm{2} \\ $$$$\mathrm{So},\:\mathrm{we}\:\mathrm{have} \\ $$$$\left({b}_{\mathrm{1}} ,\:{b}_{\mathrm{2}} \right)\:\rightarrow\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\left({b}_{\mathrm{2}} ,\:{b}_{\mathrm{3}} \right)\:\rightarrow\:\mathrm{2} \\ $$$$\left({b}_{\mathrm{1}} ,\:{b}_{\mathrm{3}} \right)\:\rightarrow\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\left({b}_{\mathrm{2}} ,\:{b}_{\mathrm{4}} \right)\:\rightarrow\:\mathrm{2} \\ $$$$\left({b}_{\mathrm{1}} ,\:{b}_{\mathrm{4}} \right)\:\rightarrow\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\left({b}_{\mathrm{3}} ,\:{b}_{\mathrm{4}} \right)\:\rightarrow\:\mathrm{2} \\ $$$$\mathrm{So},\:\mathrm{we}\:\mathrm{have}\:\mathrm{12}\:\mathrm{possibilities}. \\ $$$$\mathrm{2}\:+\:\mathrm{2}\:+\:\mathrm{2}\:+\:\mathrm{2}\:+\:\mathrm{2}\:+\:\mathrm{2}\:=\:\mathrm{12} \\ $$

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