標題:
發問:
(a) How many different numbers of 4 digits may be formed with the digits 0,1,2,5,6,7,8 if no digit is used more than once in any number. (b) How many of the numbers formed in (a) are even. (c) Find the sum of all the numbers formed in (a). 更新: (b) 420 A detailed answer for (c) is expected.
最佳解答:
First digit 6 choices : all except 0 Second digit 6 choices : all except first digit chosen Third digit 5 choices : all except first 2 digits chosen Last digit 4 choices : all except first 3 digits chosen Together (6)(6)(5)(4) = 720 Consider last digit first: Case (I) last digit is zero Last digit only 1 choice : 0 First digit 6 choices : all except zero Second digit 5 choices : all except 0 and first digit chosen Third digit 4 choices : all except the 3 digits chosen Together (1)(6)(5)(4) = 120 Case (I) last digit is non-zero Last digit only 3 choices : 2, 6, 8 First digit 5 choices : all except zero and last digit chosen Second digit 5 choices : all except first and last digit chosen Third digit 4 choices : all except the 3 digits chosen Together (3)(5)(5)(4) = 300 Case (I) + Case (II) = 420 Consider the case when zero is allowed in the first digit, there will be (7)(6)(5)(4) ways = 840 to form the numbers. Since there are 7 numbers to choose from, each number will appear 840/7 = 120 times in each of the digit positions. Now since zero is not allowed in the first digit, 120 of the 0xxx cases were removed. Therefore each of the 6 numbers will appear 120/6 = 20 times less in the 2nd, 3rd and last digits. The number of occurrences in the first digit remains the same as only 0 is disallowed. So the total is calculated as: 120(1 + 2 + 5 + 6 + 7 + 8)(1000) + 100(1 + 2 + 5 + 6 + 7 + 8)(100) + 100((1 + 2 + 5 + 6 + 7 + 8)(10) + 100(1 + 2 + 5 + 6 + 7 + 8) = 120000(29) + 10000(29) + 1000(29) + 100(29) = 3801900
其他解答:
(a) How many different numbers of 4 digitsmay be formed with the digits 0，1，2，5，6，7，8 if no digit is used more than once in any number. Sol 6*6*5*4=720 or 7*6*5*4－1*6*5*4=840-120=720 (b) How many of the numbers formed in (a) are even. (1) fourth is=0 7*6*5*1=210 (2) fourth is=2 or 6 or 8 5*5*4*3=300 210+300=510 (c) Find the sum of all the numbers formed in (a) 1+2+5+6+7+8=29 840/7=120，120/6=20 29*120*1111－29*20*111 =380190031C9A75CB3B14398
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Combination and permutation發問:
(a) How many different numbers of 4 digits may be formed with the digits 0,1,2,5,6,7,8 if no digit is used more than once in any number. (b) How many of the numbers formed in (a) are even. (c) Find the sum of all the numbers formed in (a). 更新: (b) 420 A detailed answer for (c) is expected.
最佳解答:
First digit 6 choices : all except 0 Second digit 6 choices : all except first digit chosen Third digit 5 choices : all except first 2 digits chosen Last digit 4 choices : all except first 3 digits chosen Together (6)(6)(5)(4) = 720 Consider last digit first: Case (I) last digit is zero Last digit only 1 choice : 0 First digit 6 choices : all except zero Second digit 5 choices : all except 0 and first digit chosen Third digit 4 choices : all except the 3 digits chosen Together (1)(6)(5)(4) = 120 Case (I) last digit is non-zero Last digit only 3 choices : 2, 6, 8 First digit 5 choices : all except zero and last digit chosen Second digit 5 choices : all except first and last digit chosen Third digit 4 choices : all except the 3 digits chosen Together (3)(5)(5)(4) = 300 Case (I) + Case (II) = 420 Consider the case when zero is allowed in the first digit, there will be (7)(6)(5)(4) ways = 840 to form the numbers. Since there are 7 numbers to choose from, each number will appear 840/7 = 120 times in each of the digit positions. Now since zero is not allowed in the first digit, 120 of the 0xxx cases were removed. Therefore each of the 6 numbers will appear 120/6 = 20 times less in the 2nd, 3rd and last digits. The number of occurrences in the first digit remains the same as only 0 is disallowed. So the total is calculated as: 120(1 + 2 + 5 + 6 + 7 + 8)(1000) + 100(1 + 2 + 5 + 6 + 7 + 8)(100) + 100((1 + 2 + 5 + 6 + 7 + 8)(10) + 100(1 + 2 + 5 + 6 + 7 + 8) = 120000(29) + 10000(29) + 1000(29) + 100(29) = 3801900
其他解答:
(a) How many different numbers of 4 digitsmay be formed with the digits 0，1，2，5，6，7，8 if no digit is used more than once in any number. Sol 6*6*5*4=720 or 7*6*5*4－1*6*5*4=840-120=720 (b) How many of the numbers formed in (a) are even. (1) fourth is=0 7*6*5*1=210 (2) fourth is=2 or 6 or 8 5*5*4*3=300 210+300=510 (c) Find the sum of all the numbers formed in (a) 1+2+5+6+7+8=29 840/7=120，120/6=20 29*120*1111－29*20*111 =380190031C9A75CB3B14398
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