The total number of irrational terms in the binomial expansion of $$\left(7^{1/5} - 3^{1/10}\right)^{60}$$ is

We begin with the binomial expression $$\left(7^{1/5}-3^{1/10}\right)^{60}.$$ According to the binomial theorem, which states that for any real numbers $$a$$ and $$b$$ and a non-negative integer $$n$$ we have $$\left(a+b\right)^n=\sum_{k=0}^{n}\binom{n}{k}a^{\,n-k}b^{\,k},$$ the general term of the expansion of $$\left(7^{1/5}-3^{1/10}\right)^{60}$$ is obtained by setting $$a=7^{1/5},\;b=-3^{1/10},\;n=60.$$

So the $$(k+1)\text{-th}$$ term (with $$k$$ starting from $$0$$) is

$$T_{k+1}=\binom{60}{k}\left(7^{1/5}\right)^{60-k}\left(-3^{1/10}\right)^{k}.$$

We separate the sign factor $${(-1)}^{k}$$ and write the powers of $$7$$ and $$3$$ explicitly:

$$T_{k+1}=(-1)^{k}\binom{60}{k}\;7^{\,(60-k)/5}\;3^{\,k/10}.$$

For the term $$T_{k+1}$$ to be rational, both exponents $$\dfrac{60-k}{5}$$ and $$\dfrac{k}{10}$$ must be whole numbers, because $$7^{\text{integer}}\times3^{\text{integer}}$$ is an integer (apart from the possible sign) and therefore rational.

First we impose the condition

$$\frac{k}{10}\in\mathbb{Z}\quad\Longrightarrow\quad k\text{ must be divisible by }10.$$

Within the range $$0\le k\le60,$$ the values of $$k$$ satisfying this requirement are $$k=0,\,10,\,20,\,30,\,40,\,50,\,60.$$

Next we check the second condition

$$\frac{60-k}{5}\in\mathbb{Z}\quad\Longrightarrow\quad60-k\text{ must be divisible by }5.$$

If $$k$$ is already a multiple of $$10,$$ then $$60-k$$ is automatically a multiple of $$5$$ (because subtracting a multiple of $$10$$ from $$60$$ leaves a number ending in $$0$$, which is divisible by $$5$$). Thus every one of the above $$k$$-values satisfies the second condition as well.

Therefore, exactly the following $$7$$ terms are rational:

$$k=0,\;10,\;20,\;30,\;40,\;50,\;60.$$

The total number of terms in a binomial expansion with exponent $$60$$ is $$60+1=61.$$

Hence the number of irrational terms is obtained by subtracting the $$7$$ rational terms from the total number of terms:

$$\text{Number of irrational terms}=61-7=54.$$

Hence, the correct answer is Option C.