A Basic PigeonHole Question
Allison has a box of coloured ribbons. There are 12 red ribbons, 11 white ribbons, 10 yellow ribbons, 7 blue ribbons and 4 black ribbons. What is the least number of coloured ribbons Allison must take to ensure that she will have 6 ribbons of the same colour?
The Highest number of ribbons without having any 6 of the same colour is the case as follows:
A total of
(5 x 4) + 4 = 24 ribbons
So, the above proves the “existence of a possibility” where we can choose 24 ribbons such that “no” 6 ribbons are of same colour.
Now if we have chosen (24 + 1) = 25 ribbons instead, the one more ribbon must either be red, white, yellow or blue causing a “6 ribbon of the same colour” to “always” happen in “any case/possibility”.
So, Answer = 25 Ribbons
So, 25 ribbons is the “least” amount of ribbons to “always” ensure “some 6 ribbons” of “some colour” are all of the “same colour”.
Why we need to choose 5 from each colour?
Because in “all other cases” where we choose more than 5 of any one of the colour, we already have “6 of the same colour” proving the claim in the problem with lesser ribbons.
So, we only need to construct the “Worst case scenario against” the problem statement which is done above to see the “maximum” number of ribbons which one can choose so that “no” 6 are of same colour is —-> 24 ribbons as described in the selection above.
So if we have 25 ribbons (one more than this ‘worst case scenario’), we will have “always” have 6 ribbons of the same colour.
And this is the “least” number required to force any 6 of the same colour to “exist” (could be any colour).
Meaning also that for “any” number greater than 25, we will “always have some 6 ribbons of same colour”.
But for any number of ribbons between 1 to 24, we may or may not have 6 ribbons of same colour (“not” guaranteed).
The maths problem wants the “guaranteed always” scenario with the “least number” which is “25” accordingly.
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*This idea is called the Pigeon Hole Principle (you may Google and read up some explanations which may help you understand the argument above).