VJesus12 wrote: ↑

Tue Oct 20, 2020 3:07 am

\(\rm I. \dfrac{49!}{(7^{49})^2}\)

\( \rm II. \dfrac{49!}{(7!)^2}\)

\(\rm III. \dfrac{49!}{(42!)(7!)}\)

Rank these three quantities from least to greatest.

A. \(\rm I, II, III\)
B. \(\rm I, III, II\)
C. \(\rm II, I, III\)
D. \(\rm II, III, I\)
E. \(\rm III, I, II\)

Answer: B

Solution:

We can see that the numerators are the same, so we only need to compare the denominators. Recall that for fractions with positive numerator and denominator, when the numerators are the same, the greater the denominator, the smaller the value of the fraction.

Since 7^49 > 7^7 and 7^7 > 7!, then 7^49 > 7! and hence, (7^49)^2 > (7!)^2.

Since 42! > 7!, then (42!)(7!) > (7!)(7!) = (7!)^2.

We see that (7^49)^2 = (7^2)^49 = (49)^49 has 49 factors of 49 and (42!)(7!) also has 42 + 7 = 49 factors. However, since each of the factors in (49)^49 is greater than each of the factors in (42!)(7!), then (7^49)^2 > (42!)(7!).

Since the denominators from greatest to least are (7^49)^2, (42!)(7!) and (7!)^2, the fractions from the least to greatest are 49!/(7^49)^2, 49!/(42!)(7!) and 49!/(7!)^2.

Answer: B