Question 7.

Consider two sets A={2,3,5,7,11,13}A = \left\{2, 3, 5, 7, 11, 13 \right\} and B={1,8,27}B = \left\{1, 8, 27 \right\}. Let f be a function from A to B such that for every element in B, there is at least one element a in A such that f(a)=bf(a) = b. Then, the total number of such functions f is

A
665
B
667
C
537
D
540

Question Explanation

Text Explanation

Set A={2,3,5,7,11,13} so |A|=6

Set B={1, 8, 27} so |B|=3

Without any restrictions, each element in A can map to any of the 3 elements in B. Thus, the total number of functions is: 36=7293^6=729


Excluding Functions That Miss One Element in B: If a function does not map to an element in B, there are 2 elements in B left for mapping. The total number of such functions (for each specific element not mapped) is: 26=642^6=64

Since there are 3 elements in B, the total number of such functions is: 3x64=192

Adding Back Functions That Miss Two Elements in B: If a function misses two elements in B, there is only 1 element left for mapping. The total number of such functions is: 16=11^6=1 

Since there are 3C2^3C_2 ways to choose which two elements are missed, the total number of such functions is: 3

Using the inclusion-exclusion principle, the number of functions where all elements of B are mapped by at least one element of A is:

729-192+3=540. 

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