A school selects three teachers from six teachers to support teaching in three remote areas at the same time, with one person in each place. Among them, a and B go differently, and a and C can only go together or not at the same time______ Species

A school selects three teachers from six teachers to support teaching in three remote areas at the same time, with one person in each place. Among them, a and B go differently, and a and C can only go together or not at the same time______ Species

There are two steps,
The first step is to select three teachers and divide them into two categories
In the first category, if a goes, then c c must go and B must not go. There are three different selection methods with C31 = 3
In the second category, if a does not go, then C will definitely not go, and B may or may not go. There are four different methods for selecting C43 = 4
There are 3 + 4 = 7 different selection methods
In the second step, three teachers went to three remote areas to teach, with A33 = 6,
According to the principle of step-by-step counting, the total number of different assignment schemes is 7 × 6 = 42
So the answer is: 42

A five digit number . If ABCDE satisfies a ＜ B, B ＞ C ＞ D, D ＜ E and a ＞ D, B ＞ e (such as 3720145412), then the five digit number conforms to the "sine law"______ The five digit number accords with the "sine law"

The condition is that B is the largest, D is the smallest, and a, C, e are between the minimum and maximum
When B = 9, d = 7, a, C, e can only be 8; when d = 6, a, C, e can take 7, 8, a total of 23 species; when d = 5, a, C, e can take 6, 7, 8, a total of 33 species When d = 0, a, C, e can take 1,2 8, 83 species in total;
So this situation is 1 + 23 + +83 species
When B = 8, it is 1 + 23 + +73 species, B = 7, 1 + 23 + +63 species, B = 6, 1 + 23 + +53 species, B = 5, 1 + 23 + +43 species, B = 4, 1 + 23 + 33 species, B = 3, 1 + 23 species, B = 2, 1 species
Finally, all the situations are (1 + 23 +) +83）+（1+23+… +73）+… +1=2892．
So the answer is: 2892

Repeatable permutation and combination formula The problem is that there are countless balls of three colors. If you choose three balls randomly and put them in a basket, how many possibilities will there be? The answer is 10. I want to ask what is the formula? In fact, it was thought of by the summoner playing dota. Ha ha In fact, the idea can be like this, If there are only three colors of balls, there are three situations: 1 color C31 Two colors C32 3 colors C33 C31+C21+C33=10 If we infer N kinds, what formula is it? Does anyone know?

If there is this formula, I also want to learn it. I only know the algorithm of classification. For example, let the three colors be 123, then, the classification and counting are as follows: all three colors have 123, only one color 111 222

Is there a formula for permutation and combination that can be repeated? What is the arrangement of high school learning Combination CMN = n! / m! × (n-m)! Permutation amn = n! / (n-m)! But they can't be repeated Is there a formula that can be repeated For example, a combination of 5 out of 1-9 numbers can be 11111 22225 But the above formula does not apply to this case

It doesn't need a formula, and the power is direct. For example, if 1-9 is composed of five digits, there are 9 * 9 * 9 * 9 = 9 ^ 5 cases

What's the total number of each three digit permutation? The most complete answer. Thank you For example, 123314 and so on, every three groups, what are the combined numbers? There are friends who can arrange to help list which, the more complete the better, thank you!

What is this for? I programmed it
012,013,014,015,016,017,018,019,021,023,024,025,026,027,028,029,
031,032,034,035,036,037,038,039,041,042,043,045,046,047,048,049,
051,052,053,054,056,057,058,059,061,062,063,064,065,067,068,069,
071,072,073,074,075,076,078,079,081,082,083,084,085,086,087,089,
091,092,093,094,095,096,097,098,102,103,104,105,106,107,108,109,
120,123,124,125,126,127,128,129,130,132,134,135,136,137,138,139,
140,142,143,145,146,147,148,149,150,152,153,154,156,157,158,159,
160,162,163,164,165,167,168,169,170,172,173,174,175,176,178,179,
180,182,183,184,185,186,187,189,190,192,193,194,195,196,197,198,
201,203,204,205,206,207,208,209,210,213,214,215,216,217,218,219,
230,231,234,235,236,237,238,239,240,241,243,245,246,247,248,249,
250,251,253,254,256,257,258,259,260,261,263,264,265,267,268,269,
270,271,273,274,275,276,278,279,280,281,283,284,285,286,287,289,
290,291,293,294,295,296,297,298,301,302,304,305,306,307,308,309,
310,312,314,315,316,317,318,319,320,321,324,325,326,327,328,329,
340,341,342,345,346,347,348,349,350,351,352,354,356,357,358,359,
360,361,362,364,365,367,368,369,370,371,372,374,375,376,378,379,
380,381,382,384,385,386,387,389,390,391,392,394,395,396,397,398,
401,402,403,405,406,407,408,409,410,412,413,415,416,417,418,419,
420,421,423,425,426,427,428,429,430,431,432,435,436,437,438,439,
450,451,452,453,456,457,458,459,460,461,462,463,465,467,468,469,
470,471,472,473,475,476,478,479,480,481,482,483,485,486,487,489,
490,491,492,493,495,496,497,498,501,502,503,504,506,507,508,509,
510,512,513,514,516,517,518,519,520,521,523,524,526,527,528,529,
530,531,532,534,536,537,538,539,540,541,542,543,546,547,548,549,
560,561,562,563,564,567,568,569,570,571,572,573,574,576,578,579,
580,581,582,583,584,586,587,589,590,591,592,593,594,596,597,598,
601,602,603,604,605,607,608,609,610,612,613,614,615,617,618,619,
620,621,623,624,625,627,628,629,630,631,632,634,635,637,638,639,
640,641,642,643,645,647,648,649,650,651,652,653,654,657,658,659,
670,671,672,673,674,675,678,679,680,681,682,683,684,685,687,689,
690,691,692,693,694,695,697,698,701,702,703,704,705,706,708,709,
710,712,713,714,715,716,718,719,720,721,723,724,725,726,728,729,
730,731,732,734,735,736,738,739,740,741,742,743,745,746,748,749,
750,751,752,753,754,756,758,759,760,761,762,763,764,765,768,769,
780,781,782,783,784,785,786,789,790,791,792,793,794,795,796,798,
801,802,803,804,805,806,807,809,810,812,813,814,815,816,817,819,
820,821,823,824,825,826,827,829,830,831,832,834,835,836,837,839,
840,841,842,843,845,846,847,849,850,851,852,853,854,856,857,859,
860,861,862,863,864,865,867,869,870,871,872,873,874,875,876,879,
890,891,892,893,894,895,896,897,901,902,903,904,905,906,907,908,
910,912,913,914,915,916,917,918,920,921,923,924,925,926,927,928,
930,931,932,934,935,936,937,938,940,941,942,943,945,946,947,948,
950,951,952,953,954,956,957,958,960,961,962,963,964,965,967,968,
970,971,972,973,974,975,976,978,980,981,982,983,984,985,986,987

3-digit 0-9 permutation (all results) Password box password forgotten, ha ha

From 000 to 999, there are 1000 combinations. That is, from 0 to 999
Suggestion: first think about the approximate number of 100, so as to narrow the range as much as possible, and then try from N00 to n99 one by one. The hit rate is relatively large and it is not easy to confuse. If the 100 is missing, change to another 100, which is only 1000 numbers, 10 groups

A combination of numbers from 0 to 9 How many combinations of the mantissa (i.e. the last digit) of the result number equal to 1 or 4 or 7 (again 3 or 6 or 0 or 9; and then 2 or 5 or 8) is the combination of 5 numbers randomly selected from 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9? Examples are as follows: Of mantissa 4: 0 + 0 + 1 + 1 + 2; 4 + 5 + 6 + 9 + 0; 8 + 6 + 8 + 5 + 7; 3 + 1 + 9 + 9 + 2 Of mantissa 9: 9 + 9 + 2 + 4 + 5; 0 + 1 + 2 + 3 + 3; 5 + 5 + 6 + 6 + 7; 8 + 4 + 5 + 0 + 2 Of mantissa 2: 2 + 2 + 5 + 9 + 4; 5 + 5 + 0 + 0 + 2

A total of 1452 matches, there are 20000 bytes, which can't be written down here. Please leave an address and I'll send it to you
Among them, 0 ends 88, 1 ends 89, 2 ends 90, 3-99, 4-109, 5-130, 6-152, 7-187, 8-227, 9-281
0 only:
1,2,3,5,9
1,2,3,6,8
1,2,4,5,8
1,2,4,6,7
1,3,4,5,7
1,5,7,8,9
2,3,4,5,6
2,4,7,8,9
2,5,6,8,9
3,4,6,8,9
3,5,6,7,9
4,5,6,7,8
1,2,3,7,7
1,2,4,4,9
1,2,5,6,6
1,2,5,5,7
1,2,2,6,9
1,2,2,7,8
1,1,2,7,9
1,3,4,6,6
1,3,4,4,8
1,3,3,4,9
1,3,5,5,6
1,3,3,5,8
1,3,3,6,7
1,1,3,6,9
1,1,3,7,8
1,3,8,9,9
1,4,4,5,6
1,1,4,5,9
1,1,4,6,8
1,4,7,9,9
1,4,8,8,9
1,1,5,6,7
1,5,6,9,9
1,6,7,8,8
1,6,7,7,9
1,6,6,8,9
2,3,4,4,7
2,3,3,4,8
2,2,3,4,9
2,3,3,5,7
2,2,3,5,8
2,2,3,6,7
2,3,7,9,9
2,3,8,8,9
2,2,4,5,7
2,4,6,9,9
2,5,7,8,8
2,5,7,7,9
2,6,7,7,8
2,6,6,7,9
3,4,5,9,9
3,4,7,8,8
3,4,7,7,9
3,5,6,8,8
3,5,7,7,8
3,5,5,8,9
3,6,6,7,8
3,3,7,8,9
4,5,6,6,9
4,5,5,7,9
4,4,5,8,9
4,4,6,7,9
1,1,2,3,3
1,1,2,2,4
1,1,2,8,8
1,1,4,7,7
1,1,5,5,8
2,3,3,6,6
2,4,4,5,5
2,2,4,6,6
2,2,4,4,8
2,2,5,5,6
2,5,5,9,9
2,6,6,8,8
2,2,8,9,9
3,3,4,5,5
3,3,4,4,6
3,3,6,9,9
4,5,5,8,8
4,6,6,7,7
4,4,6,8,8
4,4,7,7,8
5,5,6,7,7
5,5,6,6,8
6,8,8,9,9
7,7,8,9,9

Comparison of permutation and combination of consult numbers I have a group of numbers, such as 123 three numbers. I need to find out whether there is a number composed of these three numbers in another large group of three digit data (the position can be changed, such as 123 / 321 / 132, etc.), please consult the algorithm, thank you. I want to implement the search program on the computer Loop comparison is OK when the number of digits is small, but if I have more than 10 digits, it is too much to arrange and combine

The algorithm is as follows:
Check each character in a large string one by one
For example: 787309321890980
Let's see if it contains one of the three numbers, STR = 123 (variable position, of course)
Until the number 1 or 2 or 3 is detected, then (if 3 is detected), STR will be changed to "12"
Check whether the next number is 1 or 2, and if not, continue to look down until STR is changed to

Arrangement and combination of 10 numbers How many times can you combine from 0 to 9 If you can, tell me the combined number Of course, write as much as you can

Six permutations were selected from 10 numbers, and there were A10 (6) = 151200 methods
In the case of 0 in the first place, there are A9 (5) = 15120 methods
So there are altogether: A10 (6) - A9 (5) = 136080 methods

Number permutation and combination problem It is composed of 1,2,3,4,5,6,7, which is composed of seven digits without repetition. Find the number of seven digits of three even numbers that are not adjacent to each other

In the typical partition method, the odd number is arranged first, 4!;
Take three even numbers and five blanks, A5, 3
Multiply