Question

If they are not necessarily of different flavours?

• A. $720$
• B. $120$
• C. $126$
• D. $24$

Answer 1

Answer: (C)

$126$

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

For choosing $4$ different ice-creams the different possibilities are:

1) Choosing ice-creams of only $1$ flavour

The number of ways of selecting $1$ flavour out of $6$ is $={}^6{C}_1=6$

2) Choosing ice-creams of only $2$ flavours

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$=15\times 3=45$

3) Choosing ice-creams of $3$ flavours

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 $=20\times 3=60$

4) Choosing ice-creams of $4$ flavours

The number of ways of selecting $4$ flavours out of $6$ is$={}^6{C}_4=15$

Also different solutions for number of ice-creams of each flavour can be found out by finding number of solutions of $a+b+c+d=4$ which is ${}^3{C}_3=1$

Hence the answer $=15\times 1=15$

Hence the number of ways a customer can choose $4$ ice-creams $=6+45+60+15=126$

Hence the correct answer is $126$.