Separate disjoint cases from inclusion–exclusion

There are three identical E’s and one each of L, V and N. Thus $6!/3!=120$.

Treat EEE, L, V and N as four objects. The E’s have only one internal order, giving $4!=24$.

Fix E first and N last. The four middle letters are E, E, L and V, so their arrangements number $4!/2!=12$.

First E and last not N: if the last letter is L or V, there are $\binom21(4!/2!)=24$ arrangements of the remaining middle letters; if it is E, there are $4!=24$. This case totals 48. First not E and last N: the first letter is L or V, giving $\binom21(4!/3!)=8$. First E and last N gives 12 from (iii). The disjoint total is $48+8+12=68$.

First E gives $5!/2!=60$ arrangements of E, E, L, V, N in the remaining positions. Last N gives $5!/3!=20$ arrangements of E, E, E, L, V. Their intersection, first E and last N, was counted twice and has size 12. Hence $60+20-12=68$.