Re: [計量] JJ
※ 引述《zoo22 (zoo)》之銘言：
These problems belongs to the section of Combination and Permutation
: 3.The number of numbers that can be formed by the digits 1,2,3,4,3,2,1. the odd digits are at odd places is given by
: A) 430
: B) 215
: C) 93
: D) 36
: E) 18
: Ans: E
The numbers of way to plot odd number is at odd position is 4! / 2!2! due to
1 and 3 repeat twice. As for even number, it is 3!/2!1!
Multiply these two numbers and there comes the answer --> 6*3=18
: 4. Column A: Number of ways in which 5 differently coloured beads be
: strung on a necklace.
: Column B: 5!/2
I doubt this answer
ABCDE = BCDEA in a necklace (Circular Permutation),
therefore, the answer is 5!/5 and then divide by 2(flip the necklace),
so it shall be 4!/2....
: 7. A and B are two players of a game. Both of them have same number of coins
: before the start of the game. If a person looses the game he gives his
: opponent 1 coin and if he wins he gets 1 coin. At the end A has won 4
: games. B has 8 more coins than the nu: mber he had at the start of the
: game.Assuming there is no tie in any of the games. Find the number of
: games they played.
: Ans: 16
B-8 coins ==> it won 8+#-Games A won
A won 4, so B won 12
Sum= 16
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