Admin Control Panel

New Post | Settings | Change Layout | Edit HTML | Moderate Comments | Sign Out O level A level A A1 A2 home private tuition physics chemistry mathematics maths biology trigonometry physics H2 H1 Science Score tutor tuition tuition tutoring tuition biology economics assessment exam exams exampapers exam papers NIE JC Secondary School Singapore Education tutor teach teacher school student agency

A lvl H2 Maths: Permutations Combinations

Question from http://www.sgforums.com/forums/2297/topics/302719

Four letters from the word STATISTICS are taken to form a code-word. How many of these code words contain

i) only S and T
ii) Neither T nor C

*************************

Answer:

i) 3 letters of S and 3 letters of T

SSST type ~ 4!/3! = 4 words
SSTT type ~ 4! / (2! * 2!) = 6 words
STTT type ~ 4!/3! = 4 words

Summing up = 14 words

or you could view it as for each of the letter, you have 2 choices between S or T
which means,
2 * 2 * 2 * 2 words = 16 words
But this includes SSSS and TTTT, so 16 - 2 = 14 words


ii)
Remaining letters: S S S A I I

Num of S
= 3: then either A or I (2 * 4!/3! => 2 choices of either A or I plus number of permutations for the 4 letters)

= 2: then either with AI (4!/2!) or II (4!/{2!*2!})

= 1: then AII (4!/2! permutations for the 4 letters)

Summing up,
we have number of words = 2 * 4!/3! + 4!/2! + 4!/{2!*2!} + 4!/2!
= 8 + 12 + 6 + 12
= 38 words



Related Articles by Categories



Singapore's first free online short to
medium questions and solutions database



Related Posts with Thumbnails