A box contains blue and white ribbons. If five ribbons are choosen at random, how many ways can at least 1 blue ribbon be selected?

How many of each ribbon are in the box?

To find the number of ways to select at least 1 blue ribbon out of 5 ribbons, we need to consider two cases: selecting exactly 1 blue ribbon and selecting more than 1 blue ribbon.

Case 1: Selecting exactly 1 blue ribbon
In this case, we choose 1 blue ribbon out of the total number of blue ribbons and the remaining 4 ribbons out of the total number of white ribbons. The number of ways to choose the blue ribbon is given by the combination formula: C(n, r) = n! / (r!(n-r)!). Let's assume there are 'b' blue ribbons and 'w' white ribbons in the box.

The number of ways to choose 1 blue ribbon is C(b, 1) = b! / (1!(b-1)!) = b.
The number of ways to choose 4 white ribbons is C(w, 4) = w! / (4!(w-4)!) = w! / (4! (w-4)!).

Therefore, the total number of ways to select exactly 1 blue ribbon is b * w! / (4!(w-4)!).

Case 2: Selecting more than 1 blue ribbon
In this case, we choose 2, 3, 4, or 5 blue ribbons out of the total number of blue ribbons, and the remaining white ribbons. Again, we use the combination formula to calculate the number of ways to choose the desired number of blue ribbons and white ribbons.

The number of ways to choose 2 blue ribbons is C(b, 2) = b! / (2!(b-2)!) = b! / (2! (b-2)!).
The number of ways to choose 3 blue ribbons is C(b, 3) = b! / (3!(b-3)!) = b! / (3! (b-3)!).
The number of ways to choose 4 blue ribbons is C(b, 4) = b! / (4!(b-4)!) = b! / (4! (b-4)!).
The number of ways to choose 5 blue ribbons is C(b, 5) = b! / (5!(b-5)!) = b! / (5! (b-5)!).

For each of the above cases, the number of ways to choose the remaining white ribbons is C(w, 5 - number of blue ribbons) = w! / ((5 - number of blue ribbons)! (w - (5 - number of blue ribbons))!).

Therefore, the total number of ways to select more than 1 blue ribbon is the sum of all the combinations for choosing 2, 3, 4, or 5 blue ribbons, multiplied by the respective combinations for choosing the white ribbons:

Total number of ways = (b! / (2! (b-2)!)) * (w! / ((5-2)! (w-(5-2))!)) + (b! / (3! (b-3)!)) * (w! / ((5-3)! (w-(5-3))!))

+ (b! / (4! (b-4)!)) * (w! / ((5-4)! (w-(5-4))!)) + (b! / (5! (b-5)!)) * (w! / ((5-5)! (w-(5-5))!))

Simplifying this expression will give us the total number of ways to select more than 1 blue ribbon.

Note: The actual values of 'b' and 'w' are not provided in the question, so you will need to substitute the appropriate values in the above equations to get the final answer.