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junior college 2 | H2 Maths
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Pls help!!
Date Posted: 4 years ago
Views: 567
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I hope I didnt miss out any combinations! The answer should be 13 for part (i) and 150 for part (ii)
Date Posted:
4 years ago
But why is NAAE 1?
since N and E is fixed and the two letters between is A, there is only one way to form this 4 letter code
Ahhh thank you!!
… agree with part (i) answer, but I believe that part (ii) answer should be 270.
… for arrangement with double A,
there are still 4 different letters remaining
(counting 2 same Ns as 1 letter).
4C2 ways to select 2 letters from 4. therefore,
no. of ways = 4C2 x 4!/2! = 72
… similarly for arrangements with double N,
no. of ways = 4C2 x 4!/2! = 72
total no. of ways = 6 + 72 + 72 + 120 = 270.
… and you did remember to multiply the 5C4 for last case of all letters different . . . . . . hmmmm
… for arrangement with double A,
there are still 4 different letters remaining
(counting 2 same Ns as 1 letter).
4C2 ways to select 2 letters from 4. therefore,
no. of ways = 4C2 x 4!/2! = 72
… similarly for arrangements with double N,
no. of ways = 4C2 x 4!/2! = 72
total no. of ways = 6 + 72 + 72 + 120 = 270.
… and you did remember to multiply the 5C4 for last case of all letters different . . . . . . hmmmm
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