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How many combinations?

  1. Nov 10, 2005 #1
    A manager has Rs.1000 from which he has to give a bonus to his five employees by following a condition that every employee must get atleast Rs.50 and every employee must be given integral amount of Rupees. in how many ways can he do it?
    Since them manager has to give Rs.1000 to his employees, the amount last employee is fixed once the amount to other four are fixed. So it is the number of ways in which that amount can be given to the first four employees.
    Now the four employees as a whole should get a minimum of Rs.200. in this case there is only one condition of distributing the amount among the four employees. If the total is Rs.201, then one of the employee gets Rs.51 and thus there are 4 conditions depending on to whom the extra amount is given. Thus the total of two conditions becomes 5. When the total becomes Rs.202, there can be 4 conditions with anyone of them getting Rs.52 and 6 conditions with any two of them getting Rs.51. thus the total becomes 15 if they have to be given utmost Rs.203.
    Now I tried to find a relation between the numbers. I found that 1 in the first case is 4!/4!*0!. The 5 in next case is 5!/4!*1!. The fifteen in the next case is 6!/4!*2!. The next case also gives 35 which is again easy to verify and the formula also works as 7!/4!*3!. Thus they should be given utmost Rs.950 which happens to be the case 751st case and hence the answer should be 754!/4!*750!.
    The need is to prove. So I have to support the formula given with statements. The first case is combination to four things taken all four at a time and the four objects are four sets of Rs.50. In the next case the fifth set is a set of Rs.51. And now since the sets of Rs.50 are considered as four different sets, the combination of which person should get Rs.51 also gets verified. Now in the third case there is a one more case of Rs.52. If any combination now contains Rs.51 as well as Rs.52, the set Rs.52 will become Rs.51 to make the total sum the same and thus the number of possible combinations will also not change. But when the total sum becomes Rs.950, the need is to prove that still the number of combinations will be same. Thus it is needed to prove that the combination of high amount will be adequately replaced by multiple usage of certain small amount sets. Can anybody help how to prove it?
  2. jcsd
  3. Nov 14, 2005 #2
    ruppes is Indian currency. Rs is written before it as a standard shortcut.
  4. Nov 16, 2005 #3
    Possibly simplify?

    I'm not exactly sure how to prove your results.
    Since each employee must receive at least 50 ruppes,
    could you start with the assumption that each has 50, and change the problem to one of the number of combinations of 5 integers greater than or equal to zero. Or, to avoid counting the case when people receive 0 extra ruppes, distribute 49 ruppes to each person, and find the number of combinations of 5 numbers >=1 which total 755?

    After this simplification, you may be able to search the internet or textbooks for a more general solution which you can adapt.
  5. Nov 18, 2005 #4
    Your method is as same asmy mthod and has the same complications. the only difference is that allthe sets in my method is subtracted by 50.
    i didn't understand what you meant by searching net?
    does net provide such solutions? pls specify
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