1 1 vote 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? $20$ $35$ $41$ $21$ Combinatory isi2018-dcg combinatory + – gatecse 923 views answer comment Share Follow Print See 1 comment 1 1 comment reply txds commented Sep 20, 2019 reply Follow flag $ ^6C_1 + {^6}C_2 + {^6}C_3 = 6 + 15 +20 = 41$ 1 1 replyShare Please log in or register to add a comment.
2 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 Ashwani Kumar 2 answered Sep 20, 2019 Ashwani Kumar 2 comment Share Follow 0 reply Please log in or register to add a comment.
1 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. `JEET answered Sep 21, 2019 • edited Nov 10, 2019 by `JEET `JEET comment Share Follow 0 reply Please log in or register to add a comment.