Question
In a shop there are five types of ice-creams available. A child buys six ice-creams.
Statement - 1 : The number of different ways the child can buy the six ice-creams is $$^{10}{C_5}.$$
Statement - 2 : The number of different ways the child can buy the six ice-creams is equal to the number of different ways of arranging $$6\,A’s$$ and $$4\,B’s$$ in a row.
A.
Statement - 1 is false, Statement - 2 is true
B.
Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1
C.
Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1
D.
Statement - 1 is true, Statement - 2 is false
Answer :
Statement - 1 is false, Statement - 2 is true
Solution :
The given situation in statement - 1 is equivalent to find the non-negative integral solutions of the equation
$${x_1} + {x_2} + {x_3} + {x_4} + {x_5} = 6$$
which is co-eff. of $${x^6}$$ in the expansion of
$$\eqalign{
& {\left( {1 + x + {x^2} + {x^3} + ......\infty } \right)^5} = {\text{co-eff}}{\text{. of }}{x^6}{\text{ in }}{\left( {1 - x} \right)^{ - 5}} \cr
& = {\text{co-eff}}{\text{. of }}{x^6}{\text{ in}}\,\,1 + 5x + \frac{{5.6}}{{2!}}{x^2}...... \cr
& = \frac{{5.6.7.8.9.10}}{{6!}} = \frac{{10!}}{{6!4!}} = {\,^{10}}{C_6} \cr} $$
∴ Statement - 1 is wrong.
Number of ways of arranging $$6A’s$$ and $$4B’s$$ in a row
$$ = \frac{{10!}}{{6!4!}} = {\,^{10}}{C_6}$$ which is same as the number of ways the child can buy six ice-creams.
∴ Statement - 2 is true.