Question

One hundred identical marbles are to be distributed to three children so that each gets atlest 20 and no two get equal number. The number of ways of doing this is –

• A. 712
• B. 720
• C. 795
• D. 810

Answer 1

Answer: (C)

795

Answer 2

Let the children be A,B,C

A & B & C \end{matrix} }{\left. \ \begin{aligned} & 202159 \\ & 3941 \end{aligned} \right\} \rightarrow 19 }{\left. \ \begin{aligned} & 212257 \\ & 3940 \end{aligned} \right\} \rightarrow 18 }{.................. }{\left. \ 323335 \right\} \rightarrow 1}$$ Total number of ways = 1 + 2 + ..... + 19 – (17 + 14 + 11 + 8 + 5 + 2) = 190 – 57 = 133 By termination Permutation 133 × 3 ! = 798