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Let the boys get `x_(1),x_(2) " and " x_(3)` toys respectively. <br> `therefore x_(1)+x_(2)+x_(3)=14, x_(1), x_(2),x_(3) ge 1 " and " x_(1),x_(2) " and " x_(3)` are distinct <br> let `x_(1) lt x_(2) lt x_(3) " and " x_(2)=x_(1)+alpha, x_(3)=x_(2)+beta, " where " alpha, beta ge 1` <br> `therefore x_(1)+(x_(1)+alpha)+(x_(1)+alpha+beta)=14` <br> or `3x_(1)+2alpha+beta=14, x_(1), alpha, beta ge 1` <br> The number of solutions is equal to the coefficient of `p^(14) " in " (p^(3)+p^(6)+p^(9)..)(p^(2)+p^(4)+..)(p+p^(2)+..)` <br> = coefficient of `p^(8) " in " (1+p^(3)+p^(6))(1+p^(2)+p^(4)+p^(6)+p^(8))xx(1+p+p^(2)+..+p^(8))` <br> = coefficient of `p^(8) " in " (1+p^(2)+p^(3)+p^(4)+p^(5)+2p^(6)+p^(7)+2p^(8))xx(1+p+p^(2)+..+p^(8))` <br> `= 1+1+1+1+1+2+1+2=10` <br> Since three distinct number can be assigned to three boys in 3! ways. <br> So, corresponding to each solution, we have six ways of distribution. <br> Therefore, total numbers of ways `=10xx6=60`.**What is factorial Zero Factorial examples**

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