Since John can exhibit 1 or more of his 6 pictures, it can be done in
1(if all the pictures are displayed)
1(if only X is displayed)
1*5c1 = 5 (if X and 1 of the other 5 are displayed)
1*5c2 = 10 (if X and 2 of the other 5 are displayed)
1*5c3 = 10 (if X and 3 of the other 5 are displayed)
1*5c4 = 5(if X and 4 of the other 5 are displayed)
Total ways : 1+5+10+10+5+1 = 32(Option D)
1(if all the pictures are displayed)
1(if only X is displayed)
1*5c1 = 5 (if X and 1 of the other 5 are displayed)
1*5c2 = 10 (if X and 2 of the other 5 are displayed)
1*5c3 = 10 (if X and 3 of the other 5 are displayed)
1*5c4 = 5(if X and 4 of the other 5 are displayed)
Total ways : 1+5+10+10+5+1 = 32(Option D)
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