ISI2018-DCG-8

Question

A Pizza Shop offers $6$ different toppings, and they do not take an order without any topping. I can afford to have one pizza with a maximum of $3$ toppings. In how many ways can I order my pizza?

• A. $20$
• B. $35$
• C. $41$
• D. $21$

Comments

$ ^6C_1 +  {^6}C_2 + {^6}C_3 = 6 + 15 +20 = 41$

Answer 1

Number of ways = No of pizza with 1 topping + No of pizza with 2 topping + No of pizza with 3 topping

= $ 6C1 + 6C2 + 6C3$

= $ 6+15+20= 41$

Option C) is correct

Answer 2

Answer: $\mathbf C$

Explanation:

$\because$  pizza cannot be ordered without any topping and pizza can be ordered with at most 3 toppings.

So, only $3$ possibilities are there  $$= 6C_1 + 6C_2 + 6C_3$$

$$ = 6 + 15 + 20$$

$$ = 41$$

$\therefore \mathbf C$ is the correct option.