find number of distinguishable permutations of 2 E's, 3 J's and 4 L's
\(\frac{9!}{2!*3!*4!}= \frac{5*6*7*8*9}{2*2*3}=\frac{5*7*4*9}{1}=5*2*2*7*9=126*10=1260\)
