Select any 3 of the 7 toppings (in 7C3 ways).
Subtract those selecting in which the pair of (A,G) is definitely selected (in 1C1 ways) and any third topping is selected(in 5C1 ways)
= 7C3 - 1C1*5C1
= 30

[spoiler](E)[/spoiler] is the answer.

please explain in detail. still i could n't understand the solution

Ok, I am not sure where you're stuck so I will try to explain each part of the solution in detail.

A few concepts pertaining to this problem:
(i) The number of ways of selecting r objects from n different objects is given by
nCr. And, nCr = (n!)/[(n - r)!*(r!)]
(ii) AND means multiplication

There are 7 types of toppings from A to G.
The problem says that Al has to select 3 toppings from these 7 toppings.
In how many ways can that be done?
7C3 = (7!)/(4!)*(3!) = 35

But, there is small constraint. Al doesn't like A and G to be selected. So, from the total of 35 ways we have to exclude those ways in which A and G are both selected.
Since A and G go together, we can group then together like this: (A,G)

In how many ways can three toppings be selected that definitely includes (A,G)?
First select the group (A,G). This can be done in 1C1 ways. AND, then select any one from the remaining 5 toppings to complete the group of 3 toppings. This can be done in 5C1 ways.
So, total number of such ways = Selecting (A,G) AND Selecting any one of the remaining = 1C1*5C1 = 1*5 = 5

Required number of ways = 35 - 5 = 30

amsm25 wrote:There are 7 type of pizza toppings that Al can order put on his pizza:A,B,C,D,E,F,G. Al hates the combination of A and G. How many different combinations of 3 topping could Al order that he likes(assuming that he orders any toppings no more than once in any given combination)
1)70
2)60
3)50
4)35
5)30

AO- 5)30.

Good combinations = total possible combinations - bad combinations.

Total possible combinations:
Number of options for the first topping = 7.
Number of options for the second topping = 6.
Number of options for the third topping = 5.
To combine these options, we multiply:
7*6*5.

The product above represents the number of ways that the 3 toppings can be ARRANGED: ABC, CAB, etc.
But here we need to count the number of ways that the 3 toppings can be COMBINED.
In a combination, the ORDER of the toppings doesn't matter: ABC is the same combination as CAB.
So that we don't overcount these duplicate combinations, we must divide by the number of ways to arrange 3 elements:
3! = 3*2*1.

Thus:
The number of combinations of 3 toppings that can be formed from 7 options = (7*6*5)/(3*2*1) = 35.

But among these 35 combinations are those that include both A and G.
These bad combinations must be subtracted from the total.
Thus, the correct answer must be less than 35.

The correct answer is E.