Permutations question: book's soln to "How many diff. arrangements can be made if V is 2nd, Z never last?"

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pazzy78

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For me I;m thinking surely it should be 4 * 1 * 3 * 2 * 1

Why only 3 possibilities in the 1st box ?

The only one it can;t be is V , as that must be 2nd ...
 
I get 18 (= 4! - 3!) too, but I don't understand the explanation in the book.
 
pazzy78 said:
For me I;m thinking surely it should be 4 * 1 * 3 * 2 * 1

Why only 3 possibilities in the 1st box ?

The only one it can;t be is V , as that must be 2nd ...
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Because you've already filled the 2nd and 5th boxes, so there are 3 choices left.
 
pazzy78 said:
No , Z can never be last, V is the only restricted one to 2nd place here, so in theory W,X,Y and Z can be 1st ....
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No, you're missing the point. They said, "Start with the choice that has most restrictions, and choose a possible letter each time."

So you first "choose" one for position 2, which has to be V, leaving 4 letters. Then choose one for position 5, which has to be W, X, or Y, That leaves 3 more letters. Suppose you chose Y; then there are W, X, and Z left.

Now choose one of those 3 form position 1, then one of the remaining 2 for position 3, and the last one for position 4.

The counts to be multiplied are the number of choices available at that point in the process, not the total number of possibilities for that place.
 
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