# My question is:How many 15 letter arrangement's of 5 A's , 5 B's , 5 C's have no A's in the first 5 letters , no B's in next 5 letters , and no C's in the last 5 letters

Aman Bansal
592 Points
12 years ago

dear Anchit,

at first it appears that first 5 letters are to be chosen from Bs and Cs, so there are 10 * 9 * 8 * 7 * 6 ways to satisfy the condition that there would be no As in first 5. but then next 5 is dependent on what happens in first 5 letter slots, so its "more complicated" i get that then it says "suppose that we only use 5 Bs in the first 5 places. there are 5! ways to do this. Then all C's must go in the 2nd 5 places and all A's must go in last 5 places. there are a total of (5!)^3 choices when we use only Bs in the first 5 places.. if we use 4 Bs and 1 C in the first 5 places, there are (5 1) [written vertically so its a parenthese with 5 on top of the one] ways they can be chosen..." then it goes on to explain all the other scenarios. what i don't get is, why are there 5! ways to "use only Bs in the first 5 places? isn't BBBBB the same as BBBBB? so shouldn't it be 5C5... not 5!? because if they're going to distinguish between BBBBB, then why are they suddenly doing 5 choose 1 ways in the "if we use 4 Bs and 1 C in the first 5 places?" unless the vertical parentheses is the 5 P 1..

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