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12C3*9C3*.... is akin to doing n*(n-1)*(n-2) which takes into account an order.... and in these cases you divide by appropriate x! when you want to disregard the order or when there are x of same type. right ?

No. of ways to form 1st team: 12*11*10= 1320 No. of ways to form 2nd team: 9*8*7=504 No. of ways to form 3rd team: 6*5*4=120 No. of ways to form 4th team: 3*2*1=6

No. of ways to form 1st team: 12*11*10= 1320 No. of ways to form 2nd team: 9*8*7=504 No. of ways to form 3rd team: 6*5*4=120 No. of ways to form 4th team: 3*2*1=6

therefore total: 1950 (ans)

Whats the OA/OE?

this is incorrect you are taking permutaitons. Here order is not important. No. of ways to form 1st team: 12*11*10= 1320 it should be 12c3

take another example if there 3 persons and need to form team of 3 person. how many ways you can do it. PEROSNS A B C only one team can be formed .. ABC here order doesn't matter ABC=BCA=CBA=.. 3C3

If question asks how do you arange 3 persons in 3 seats. then 3!
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