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Consider the number N=2012. Find the number of cyphers at the end of ""^(N)C_(N//2). |
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Answer» SOLUTION :`""^(N)C_(N//2)=""^(2012)C_(1006)=((2012)!)/((1006)!(1006)!)` NUMBER of 5's in numerator =402+80+16+3=501 number of 5's in denominator`=2[201+40+8+1]=500` `:.` number of cyphers in `""^(N)C_(N//2)` is 1 |
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