Question

If they contain only $2$ or $3$ different flavours?

• A. $105$
• B. $115$
• C. $120$
• D. $135$

Answer 1

The correct option is A $105$

Number of positive solutions for $x_1+x_2+x_3+....x_n=k$ is ${}^{k-1}{C}_{n-1}$

1) The number of ways a customer can choose $4$ ice-creams from $2$ different flavours can be calculated as

The number of ways of selecting $2$ flavours out of $6$ is $={}^6{C}_2=15$

Also different solutions for number of ice-creams of each flavour can be found out by finding number of solutions of $a+b=4$ which is ${}^3{C}_1=3$

Hence the answer $=15\times 3=45$

2) The number of ways a customer can choose $4$ ice-creams from $3$ different flavours can be calculated as

The number of ways of selecting $3$ flavours out of $6$ is $={}^6{C}_3=20$

Also different solutions for number of ice-creams of each flavour can be found out by finding number of solutions of $a+b+c=4$ which is ${}^3{C}_2=3$

Hence the answer$=20\times 3=60$

Hence the number of ways a customer can choose $4$ ice-creams from $2$ or $3$ different flavours $=45+60=105$