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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?

  1. $20$
  2. $35$
  3. $41$
  4. $21$
in Combinatory by Boss (16.8k points)
recategorized by | 25 views
+1
$ ^6C_1 +  {^6}C_2 + {^6}C_3 = 6 + 15 +20 = 41$

2 Answers

+2 votes

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

by Boss (15.6k points)
+1 vote

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.

by Boss (13.2k points)
edited by
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