×

# Permutation & Combination

How many four letter words can be formed using the letters of the word 'INEFFECTIVE' so that (i)3 alike letters and one distinct letter (ii)2 alike letters of one kind & 2 distinct letters (iii)all distinct letters are there

Note by Dharmik Panchal
4 years, 2 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

I) The only possible three repeated letters are Es, because only E is repeated thrice. The choice of the fourth letter is from I, N, F, C, T and V (6 choices).

Total number of permutations of these is hence 6C1 x (4!) / (3!) = 24.

II) There are 3 possibilities for the 2 alike letters: either II, EE or FF. If you choose I as your repeated letter, any 2 letters from N, E, F, C, T and V will be your choices for the two distinct letters. If you choose E as your repeated letter, any two letters from N, I, F, C, T and V will be your choices for the two distinct letters. If you choose F as your repeated letter, any two letters from E, N, I , C, T and V will be your choices for the two distinct letters.

Total number of permutations is hence 3C1 x 6C2 x (4!) / (2!) = 540

III) All your four distinct letters will be from I, N, E, F, C, T and V.

Total number of permutations is hence 6P4 = 360

- 4 years, 2 months ago