The number of sequences in which 7 players can throw a ball,so that the youngest player may not be the last ball is :

Question

a) $4000$ b) $2160$ c)$4320$ d) $5300$

MyApproach:

The first 6 players can throw a ball in 7! ways and the last player can throw in 6 ways.

Hence,Ans $7!$ . $6$

Can Anyone give me the Hint why I am wrong?

Not following your calculation. Hint: the answer is the total number of orderings (with no constraint) less the number of orderings in which the youngest player goes last. — lulu, Oct 01 '15 at 15:46
As with a lot of these problems, it is easier to do the opposite. The number of orderings in which the youngest goes last is clearly $6!$ as you can order the others however you like. The unconstrained number is $7!$. Hence $7!-6!=6^*(6!)$. Now that does resemble your answer, but I could not follow your reasoning. — lulu, Oct 01 '15 at 15:52

score 3 · Accepted Answer · answered Oct 01 '15 at 16:19

If you want to get the answer $6(6!)$ by multiplication principle (as opposed to using the complement which lulu does above and of course works perfectly well also), first pick the person who will throw last (6 ways), then order the remaining 6 people for the first 6 throws (6! ways).

The number of sequences in which 7 players can throw a ball,so that the youngest player may not be the last ball is :

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