Let A={x1,x2,x3,x4,x5,x6},B={y1,y2,y3,y4,y5,y6}.Then the number of one-one mappings from A to B such that f(x i ) not equal to y i i=1,2,3,4,5,6 is

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Ravi Jhunjhunwala · Answer

n-1 ways... thats the number of combinations possible... please verify.......as we take 2 elements one arrangement possible, 3 elements 2 arrangements possible

Ravi · Answer

For 6 terms....its 6 factorial ((1/2!)-(1/3!)+..... . Thus the answer 265... Therefore there is 265 ways in which two sets can be mapped

Ravi Jhunjhunwala · Answer

both of the above comments are by me only.

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Let A={x1,x2,x3,x4,x5,x6},B={y1,y2,y3,y4,y5,y6}.Then the number of one-one mappings from A to B such that f(x i ) not equal to y i i=1,2,3,4,5,6 is

Let A={x1,x2,x3,x4,x5,x6},B={y1,y2,y3,y4,y5,y6}.Then the number of one-one mappings from A to B such that f(x_i) not equal to y_i i=1,2,3,4,5,6 is

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Let A={x1,x2,x3,x4,x5,x6},B={y1,y2,y3,y4,y5,y6}.Then the number of one-one mappings from A to B such that f(x i ) not equal to y i i=1,2,3,4,5,6 is

Let A={x1,x2,x3,x4,x5,x6},B={y1,y2,y3,y4,y5,y6}.Then the number of one-one mappings from A to B such that f(xi ) not equal to yi i=1,2,3,4,5,6 is

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Let A={x1,x2,x3,x4,x5,x6},B={y1,y2,y3,y4,y5,y6}.Then the number of one-one mappings from A to B such that f(x_i) not equal to y_i i=1,2,3,4,5,6 is