You can order a pizza with up to 8 possible toppings. Which is greater? The number of possible pizzas you can get with only 3 toppings or the number of possible pizzas you can get with 5 toppings?
1st case = \(\large{^8C_3 \times 3!}\) 2nd case = \(\large{^8C_5\times 5!}\) Since, \(\large{^nC_r = ^nC_{n-r}}\) Thus: \[\large{^8C_3 = ^8C_5}\] Thus: 2nd case > 1st case
I have assumed these things: 1. All toppings are different. 2. When you say number of toppings on pizza, you mean the `exact` number of toppings.
3. Complete pizza = pizza base + topping(s) (with order of toppings important)
4. Different order of toppings mean different complete pizza
Why would the order of the toppings be important? :3
I know - "many assumptions" but that's what I can think of now.
Well I have not eaten many pizzas, so I am leaving that to you all
If order is not important then both have the same number
By the way, vishweshshrimali5 there's an easier way of writing \[\Large ^nC_r\cdot r!\] And that's \[\Large ^nP_r\] ^.^
Yeah I realised that while typing that stuff but left it as it gave the same answers :) But thanks.
I had not considered that the order in which the toppings are placed matters. So yes, you are right haha.
Would it matter, though? Does the pizza look like some sort edible archery target? haha
It was just an amusing "trick" question I came across and thought it was fun but all the smart people came and knew I couldn't slide it past you guys haha.
It worked (I guess) :D !! So, .... Kainui.... when are you going to offer me and Peter the pizzas we prepared
I'd love a slice :3
*number of pizzas we both found
brb lol
Hey thats not a pizza :P
who wants the ones with 5 toppings and who wants the ones with 3 toppings?
I want the one pizza with all 8 toppings.
Same here ^^^^^^^
How about I give you 1 pizza with 0 toppings, it's pretty much the same.
:/
lol
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