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We have test that is being held in 3 rooms A B C , and we have 4 teaches P Q R S. Each teacher has to be supervision in one room. The condition is that one room will have 2 teaches as supervision and no room can remain without supervision.

We are to find the number of ways how to divide teachers inside room.

If we didnt have condition. We would just divide 4 teachers into 3 rooms which is basic combination =

$\binom{n }{ k}$ = $\binom{4}{3}$ = $\binom{4!}{(4-3)!*3!}$ = 4

But how to solve it with this condition? I really appreciate all explanation and help! Thanks

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since the order should matter here , the amount od combinations dividing 4 teachers into 3 room should be 4 * 3 * 2

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  • $\begingroup$ Imagine that the teachers are pigeons? $\endgroup$ – hypergeometric Nov 17 '16 at 16:28
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There are three ways to choose the room to which two teachers are sent. There are $\binom{4}{2}$ ways to choose two of the four teachers to assign to that room. There are $2!$ ways to assign the remaining teachers to the remaining rooms. Thus, the number of ways of assigning the teachers to the rooms so that exactly one room is assigned two teachers is $$\binom{3}{1}\binom{4}{2} \cdot 2!$$

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First select three teachers out of given 4. Then send them in the three classes, one in each class, the last teacher can then be sent in any of the three classes.

The mathematical equivalent of the above procedure would be:

[(4 C 3) × (3!) × 3]/2 = 36. (Ans.)

The division by two since you can perform the two given actions (sending three teachers first and then sending the last teacher) in any order.

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  • $\begingroup$ the answer should be 36 $\endgroup$ – trolkura Nov 17 '16 at 14:45
  • $\begingroup$ as i wrote , the correct answer to this should be 36. This is from solved exercises from my school. Answer is 36 but i cannot find out why. $\endgroup$ – trolkura Nov 17 '16 at 14:52
  • $\begingroup$ Its correct now...corrected couple of minor errors. $\endgroup$ – SirXYZ Nov 17 '16 at 14:54
  • $\begingroup$ Why did you use 4 C 3 ? By sending 3 teachers into 3 rooms its 3! since order does matter here , and then the last one can be send in any of those room so * 3 but how did u got 4 C 3? $\endgroup$ – trolkura Nov 17 '16 at 15:01
  • $\begingroup$ (4 C 3) for selecting three teachers out of given 4. $\endgroup$ – SirXYZ Nov 17 '16 at 15:02

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