**Arrangement of Tokens in 2 identical Boxes**

There are n number of tokens marked 1,2,3,4,..........n respectively. |

One has to pick up 2 tokens and put them in 2 boxes, one in each. |

None of the boxes should remain empty nor have more than 1 token. |

Both the boxes are identical. |

In how many ways, 2 tokens can be allocated to both boxes out of n available tokens ? Ans : C(n,2) |

What is the minimum sum of 2 tokens so allocated ? Ans - 1+2=3 |

What is the maximum sum of 2 tokens so allocated ? Ans:- n+(n-1) = 2n-1 |

What is the average sum of 2 tokens so allocated ? xav=[3+(2n-1)] /2 = n+1 |

In how many ways, the 2 tokens can be arranged so as to get the average sum? Ans:-quotient (n/2) |

If X is the sum of 2 tokens such that X=xav ±x, then no. of arrangements to get X is same for xav+x & xav-x. |

If n is odd, no. of arrangements to get a number equal to xav ±1=No. of arrangements to get xav |

If n is even, no. of arrangements to get a number equal to xav ±1=(No. of arrangements to get xav) -1. |

Example-- No. of tokens 1,2,3,.......8. One token in each of 2 boxes. Minimum no. that can be put--3, maximum no. that can be put--15, Average no. --09 |

Token with sum-3(xav-6)
sum-4(xav-5)
sum-5(xav-4) sum-6(xav-3)
sum-7(xav-2)
sum-8(xav-1)
sum-9 (xav=9+x=0) |

1+2
1+3
1+4
1+5
1+6
1+7
1+8 2+3 2+4 2+5 2+6 2+7 3+4 3+5 3+6 4+5 |

sum-10(xav+1) sum-11(xav+2)
sum-12 (xav+3)
sum-13(xav+4) sum-14(xav+5)
sum-15(xav+6) |

2+8
3+8
4+8
5+8
6+8
7+8 3+7 4+7 5+7 6+7 4+6 5+6 |