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junior college 1 | H1 Maths
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Hi, my senior needs help with this question part (ii) (a.). Thank you :)
①Repetitions allowed
② digit 5 exactly once
③ letter H exactly twice
First 3 characters are digits. Last 3 characters are letters.
So we can have :
①5 _ _ H H _
②5 _ _ H _ H
③5 _ _ _ H H
④_ 5 _ H H _
⑤_ 5 _ H _ H
⑥_ 5 _ _ H H
⑦_ _ 5 H H _
⑧_ _ 5 H _ H
⑨_ _ 5 _ H H
This is basically saying :
Number of ways to arrange the 5 and H
= 3C1 x 3C2
= 3 x 3
= 9
So for the blanks, we cannot pick 5 or H already.
For numbers we can only choose from 12346 (total 5 choices)
For letters we can only choose from ABCDEFG (total 7 choices)
For each of the 9 arrangements above,
Number of ways to pick first blank digit
= 5
Number of ways to pick second blank digit
= 5
Number of ways to pick the blank letter
= 7
Total number of ways = 9 x 5 x 5 x 7 = 2565
Number of ways = 6 x 6 x 6 x 8 x 8 x 8
= 110592
Required probability = 2565/110592
= 95/4096
9 x 5 x 5 x 7 = 1575 instead of 2565
(Keyed in 9 x 5 x 57 by mistake in calculator)
Required probability = 1575/110592
= 175/12288
See 1 Answer
Need to analyze a bit
These are the answers
Use box method Orelse exam don’t know how to do
Likewise, for i), it would be 8C1 x 7C1 x 6C1 for the letters instead of 7C1 x 6C1 x 5C1