# What is the probability that all 3 will be consonants?

This is mainly a sanity check, because it wouldn't be the first time I discovered this textbook is wrong about something. See question #3 here:

I've tried calculating by hand $$\frac{(21!/(3!(21-3)!) }{ (26!/(3!(26-3)!))}$$, punching it carefully into my calculator, and checking against Wolfram Alpha, and I always get 51%.

How did this textbook come up with 82%? Is the formulation of C(21, 3)/C(26, 3) correct or incorrect?

EDIT:

Lookie here, I just wrote an empirical test using python and the results are closer to my answer:

• There is something weird indeed in the book. It states that ${21 \choose 3}$ would be $54~264$, but actually it's $1~330$. A HUGE difference ... hmmm ... $\frac{ { 21 \choose 3} }{ { 26 \choose 3 } }$ is indeed approximately $0.511~538$. Perhaps the maker has made a typo, because$${21 \choose 6} = 54~264 \quad \text{and} \quad {26 \choose 5} = 65~780$$ – Matti P. Feb 14 at 10:07
• Yeah that's weird... I even just wrote a little test to try this in an empirical way, and it's definitely about 51%. – user1952534 Feb 14 at 10:20
• So my answer is: You are correct, the book is wrong. – Matti P. Feb 14 at 10:54