Question

How many strings with seven or more characters can be formed from the letters in EVERGREEN?

Short answer

Therefore, there are\(19635\)different strings formed.

Step-by-step solution

Step 1: Definitions

Sum rule

If an event can occur either in m ways or in n ways(non-overlapping), the number of ways the event can occur is then the addition of m and n.

Step 2: Distribution of letters

Distributing n distinguishable objects into k distinguishable boxes such that \({n_i}\)objects are placed in box i (\(i = 1,2,3,4,5\))

The word EVERGREEN contains\(4Es,1V,2Rs,1G,1N\). We are only interested in words that have seven or more characters.

There are nine letters\(4Es,1V,2Rs,1G,1N\)

\(\frac{{n!}}{{{n_1}!{n_2}!......{n_k}!}} = \frac{{9!}}{{4!1!2!1!1!}}\)

Now for eight letters

\(4E,1V,2R,1G \Rightarrow \frac{{n!}}{{{n_1}!{n_2}!......{n_k}!}} = \frac{{8!}}{{4!1!2!1!}} = 840\)

\(4E,1V,2R,1N \Rightarrow \frac{{n!}}{{{n_1}!{n_2}!......{n_k}!}} = \frac{{8!}}{{4!1!2!1!}} = 840\)

\(4E,2R,1G,1N \Rightarrow \frac{{n!}}{{{n_1}!{n_2}!......{n_k}!}} = \frac{{8!}}{{4!1!2!1!1!}} = 840\)

\(4E,1V,1R,1G,1N \Rightarrow \frac{{n!}}{{{n_1}!{n_2}!......{n_k}!}} = \frac{{8!}}{{4!1!1!1!1!}} = 1680\)

\(3E,1V,2R,1G,1N \Rightarrow \frac{{n!}}{{{n_1}!{n_2}!......{n_k}!}} = \frac{{8!}}{{3!1!2!1!1!}} = 3360\)

Step 3: 7 letters

\(4E,1V,2R \Rightarrow \frac{{n!}}{{{n_1}!{n_2}!......{n_k}!}} = \frac{{7!}}{{4!1!2!}} = 105\)

\(4E,2R,1G \Rightarrow \frac{{n!}}{{{n_1}!{n_2}!......{n_k}!}} = \frac{{7!}}{{4!1!2!}} = 105\)

\(4E,2R,1N \Rightarrow \frac{{n!}}{{{n_1}!{n_2}!......{n_k}!}} = \frac{{7!}}{{4!1!2!}} = 105\)

\(4E,1V,1R,1N \Rightarrow \frac{{n!}}{{{n_1}!{n_2}!......{n_k}!}} = \frac{{7!}}{{4!1!1!1!}} = 210\)

\(4E,1V,1R,1G \Rightarrow \frac{{n!}}{{{n_1}!{n_2}!......{n_k}!}} = \frac{{7!}}{{4!1!1!1!}} = 210\)

\(4E,1R,1N,1G \Rightarrow \frac{{n!}}{{{n_1}!{n_2}!......{n_k}!}} = \frac{{7!}}{{4!1!1!1!}} = 210\)

\(4E,1V,1N,1G \Rightarrow \frac{{n!}}{{{n_1}!{n_2}!......{n_k}!}} = \frac{{7!}}{{4!1!1!1!}} = 210\)

\(3E,1V,2R,1G \Rightarrow \frac{{n!}}{{{n_1}!{n_2}!......{n_k}!}} = \frac{{7!}}{{3!1!2!1!}} = 420\)

\(3E,1V,2R,1N \Rightarrow \frac{{n!}}{{{n_1}!{n_2}!......{n_k}!}} = \frac{{7!}}{{3!1!2!1!}} = 420\)

\(3E,2R,1N,1G \Rightarrow \frac{{n!}}{{{n_1}!{n_2}!......{n_k}!}} = \frac{{7!}}{{3!1!2!1!}} = 420\)

\(3E,1V,1R,1N,1G \Rightarrow \frac{{n!}}{{{n_1}!{n_2}!......{n_k}!}} = \frac{{7!}}{{3!1!1!1!1!}} = 840\)

\(2E,1V,2R,1N,1G \Rightarrow \frac{{n!}}{{{n_1}!{n_2}!......{n_k}!}} = \frac{{7!}}{{3!1!2!1!}} = 1260\)

Step 4: Total

Use the sum rule:

\(\begin{array}{l}7560 + (840 + 840 + 840 + 1680 + 3360) + \\(105 + 105 + 105 + 210 + 210 + 210 + 420 + 420 + 420 + 840 + 1260)\\ = 7560 + 7560 + 4515\\ = 19635\end{array}\)

Therefore, there are\(19635\)different strings formed.