1. How many distinct permutations can be made from the letters of the word"EXCELLENT"?
2.A family of 5 has dinner at a round table. Find the number of ways they can be seated if
a)there is no restriction
b)the parents must be seated together
c)the parents must not be seated together
3. Susan decorates her Christmas tree using a series of light bulbs. She has 5red, 4 yellow and 2 blue bulbs available.
ai)if she uses all the light bulbs, how many different arrangements can she make?
aii) in (i) , how many of the arrangements will the two blue bulbs be separated?
b)if she only uses 10 of the bulbs, how many different arrangements can she make?
4. Th prime minsters A,B,C,D,E,F and G of 7 countries will address at a summit meeting.
a)find the number of arrangements that can be made so that
i)A will speak before C.
ii)A will speakbefore C and C before E
b)In how many of those ways in (aii) will C speak immediately after A?
5. If 4 letters are selected from the 7 letters of the word"WINDOWS", calculate the number of possible
a)combinations,
b)arrangements.
6. There are 3 letter-boxes marked P,Q and R and 8 letters marked A, B, C, D, E, F, G and H.
a)find the number of ways in which the 8 letters can be placed in the boxes if
i)there is no restriction,
ii)4 of them are in box P, 3 of them in box Q and 1 of them in box R
math s.5問題!!!10點!!!!
2013-01-07 12:45 am
回答 (1)
2013-01-07 6:21 am
✔ 最佳答案
1. How many distinct permutations can be made from the letters of the word"EXCELLENT"?The number of distinct permutations can be made= 9! ÷ 2! ÷ 3! i.e., There's 2 letters'L', 3 letters'E'
= 30240//2. A family of 5 has dinner at a round table. Find the number of ways they can be seated if
(a) there is no restrictionThe number of ways
= 5!
= 120//(b)the parents must be seated togetherThe number of ways
= 4! × 2 --here we treat the two parents as one
= 48//(c)the parents must not be seated togetherThe number of ways
= 120 - 48 --just the difference betwwen answers in (a) and (b)
= 72//3.Susan decorates her Christmas tree using a series of light bulbs. She has 5red, 4 yellow and 2 blue bulbs available.
(a)(i)if she uses all the light bulbs, how many different arrangements can she make?The number arrangements
= 11C5 × 6C4
= 6930//(a)(ii)in (i) , how many of the arrangements will the two blue bulbs be separated?The number of arrangements
= 6930 - 10C5 × 5C4
= 5670//(b)if she only uses 10 of the bulbs, how many different arrangements can she make?The number of arrangements
= 10C5 × 5C4 + 10C4 × 6C4 + 10C5 × 5C3
= 6930//4.Th prime minsters A,B,C,D,E,F and G of 7 countries will address at a summit meeting.
a)find the number of arrangements that can be made so that
i)A will speak before C.
The number of arrangements
= 7! ÷ 2!
= 2520//ii)A will speakbefore C and C before E
The number of arrangements
= 7! ÷ 3!
= 840//
b)In how many of those ways in (aii) will C speak immediately after A?
The number of ways
= 6! ÷ 2!
= 360//
5. If 4 letters are selected from the 7 letters of the word"WINDOWS", calculate the number of possible
a)combinations,
The number of combinations
= 5C4 + 5C3 + 5C2
= 25//
b)arrangements.
The number of arrangements
= 5P4 + 5P3 × 4C1 + 5P2 × 4C2
= 480//
6a)find the number of ways in which the 8 letters can be placed in the boxes if
i)there is no restriction,
The number of ways
= 3^8
= 6561//
ii)4 of them are in box P, 3 of them in box Q and 1 of them in box R
The number of ways
= 8C4 × 4C3
= 280//
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