How many 4-letter selections from CROCODILE have different numbers of C and O letters?

Key takeaway: This is a neat counting problem with repeated letters. The clean way is to count the opposite case first: selections where the number of C’s equals the number of O’s, then subtract from the total.

Treat the two C’s and two O’s in CROCODILE as available letter positions.

1) Count all 4-letter selections

There are 9 letter positions, so the total number of ways to choose 4 is

$\binom{9}{4}=126.$

2) Count selections where the number of C’s equals the number of O’s

Let the common number be $k$.

• Case $k=0$: choose all 4 letters from the other 5 letters. $\binom{5}{4}=5$

• Case $k=1$: choose 1 of the 2 C’s, 1 of the 2 O’s, and 2 of the other 5 letters. $\binom{2}{1}\binom{2}{1}\binom{5}{2}=2\cdot2\cdot10=40$

• Case $k=2$: choose both C’s and both O’s. $\binom{2}{2}\binom{2}{2}=1$

So the number with equal C’s and O’s is

$5+40+1=46.$

3) Subtract

Therefore, the number of selections in which the number of C’s is not the same as the number of O’s is

$126-46=80.$

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Pitfalls the pros know 👇 A common slip is to count only the four chosen letters without respecting the repeated C’s and O’s. Another easy mistake is forgetting the $k=2$ case, where both C’s and both O’s are selected. If the letters are treated as indistinguishable, the count changes, so you need to match the wording used in the question.

What if the problem changes? If the question were changed to “the number of C’s is the same as the number of O’s,” the answer would be the complement, $46$. If the selection size changed from 4 to 5, you would repeat the same strategy: split into cases by the common number of C’s and O’s, then subtract from the total.

Tags: combinations, repeated letters, complement counting