How many 5-letter selections can be made from SEVENTEEN with specified E's and N's?

Key concept: This is a selection problem, not an arrangement problem. So order does not matter. The repeated letters still matter, but you count combinations of letter types rather than permutations.

Step by step

The word SEVENTEEN contains:

• 4 E's
• 2 N's
• 1 S, 1 V, 1 T

We are choosing 5 letters.

(iii) Exactly 2 E's and exactly 2 N's

If 2 E's and 2 N's are already fixed, there is 1 more letter to choose from S, V, T.

So the number of possible selections is

$3$

(iv) At least 2 E's

We count by cases.

Case 1: exactly 2 E's

Choose 3 more letters from N, N, S, V, T.

Possible selections:

• 0 N's: choose 3 from S, V, T → 1 way
• 1 N: choose 2 from S, V, T → 3 ways
• 2 N's: choose 1 from S, V, T → 3 ways

Total for exactly 2 E's:

$1+3+3=7$

Case 2: exactly 3 E's

Choose 2 more letters from N, N, S, V, T.

Possible selections:

• 0 N's: choose 2 from S, V, T → 3 ways
• 1 N: choose 1 from S, V, T → 3 ways
• 2 N's: choose none from S, V, T → 1 way

Total for exactly 3 E's:

$3+3+1=7$

Case 3: exactly 4 E's

Then choose 1 more letter from N, S, V, or T.

That gives

$4$

Add the cases:

$7+7+4=18$

Answers

• (iii) 3
• (iv) 18

Pitfall alert

The usual trap is forgetting that the letters are being selected, not arranged. So "E then N then S" is the same selection as "S, E, N." Another easy mistake is to ignore that N appears twice, which affects the counting in the middle cases.

Try different conditions

If the question changed to "exactly 2 E's and exactly 1 N," you would still use the same case-by-case logic, but the leftover letters would be chosen from S, V, T and one additional letter. If the total number of chosen letters changed from 5 to 6, the balance of E's and N's would shift and the cases would need to be rebuilt from scratch.

Further reading

combinations, multiset selection, casework