In how many ways can you sit 10 class eleven students and 14 class twelve students together in a row if no two class eleven students may be adjacent?
I have tried it this way:
Since no two students from class eleven can sit together, we can make a 10 pairs of students by taking 1 from class eleven and 1 from class twelve. If we treat those 10 pairs as 10 individual students, we will have 14 individual students to arrange, taking the remaining 4 of class twelve students. As a result we will have 14! permutation.
The answer in text is given $^{15}P_{10} \times 14!$
Do we have to take 10 students from 15 slots in total because if we count each slot at the right of each class eleven students, there are 10 slots, 4 extra slots from the non paired class twelve students, and 1 extra slot because a class twelve student can also sit at the left of the first class eleven student in the row? Is this the reason we take 10 out of 15 instead of 10 out of 14?
Best Answer
First arrange grade $12$ learners in $14!$ ways
Then it will be $15$ spaces, from before first learner to after last learner
$(1)\text{GR}12,(2)\text{GR}12,(3)\text{GR}12,(4)\text{GR}12,(5)\text{GR}12,(6)\text{GR}12,(7)\text{GR}12,(8)\text{GR}12,(9)\text{GR}12,(10)\text{GR}12,(11)\text{GR}12,(12)\text{GR}12,(13)\text{GR}12,(14)\text{GR}12,(15)$
it can be $^{15}P_{10}$ diffrent arrangements for grade $11$ learners
total number of arrangements is $^{15}P_{10} \times 14!$