The number of values in set of values of r for which 23Cr+2⋅23Cr+1+23Cr+2≥25C15 is ______.
23Cr+2⋅23Cr+1+23Cr+2=24Cr+1+24Cr+2=25Cr+2≥25C15Therefore, (r + 2) can be 10, 11,12 ,13,14 and 15.So there are 6 elements.