In how many ways could these things been paired so that no one takes their own!?

Question

When the three Adams children went to school one morning,they each took one of the others' lunches,In how many ways could the three children have been paired with the lunches so that no one took his or her own?

(A) One
(B) Two
(C) Three
(D) Six
(E) Eight

Answer 1

let the chidren are A,B,C and their lunch are a,b,c
then the sequence may be if they dont take their own lunch
i) Ab, Bc, Cb
ii) Ac, Ba, Cb
so the ways are only 2.

Answer 2

6

Answer 3

Let us consider child A. He can only take either B's or C's lunch. Thus, A has 2 possible choices.

Now consider the child whose lunch was not taken by A. The only remaining choices are A's lunch or his own lunch. Since he cannot take his own lunch, then he has only one possible choice.

Now since that child's lunch wasn't taken yet, then it is the only remaining choice for the person whose lunch was taken by A. He has only one possible choice.

By the Fundamental Principle of Counting:
2 x 1 x 1 = 2 possible ways.

the answer is (B) two.

^_^

Answer 4

That would be (B) Two

(child1, lunch2), (child2, lunch3), (child3, lunch1)

(child1, lunch3), (child2, lunch1), (child3, lunch2)