Question

There are two different blue toys three different green toys and four different red toys in how many ways one can choose three toys such that there is at least one green toy and one blue toy in the chosen 3

Answer 1

Here there are 3 blue toys , 2 green toys & 4 red toys

So total number of toys are 9.

Now we have to select 3 toys of which there must be at least 1 blue & 1 green toy.

Case 1.

There is 1 blue, 1 green & 1 red toy

No of selections = 3C1 × 2C1 × 4C1 = 3 × 2 × 4= 24

Case 2.

There are 2 blue & 1 green toy

No of selections = 3C2 × 2C1 × 4C0= 3 × 2 × 1 = 6

Case 3.

There is 1 blue toy & 2 green toys

No of selections = 3C1 ×2C2 × 4C0= 3 ×1 ×1 = 3

Hence the total number of ways to select three toys of which at least 1 is blue & 1 is green are 24 + 6 + 3 = 33

One can choose three toys in 33 ways such that there is at least one green toy and one blue toy in the chosen 3.