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junior college 1 | H2 Maths
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Hi can someone help me with qn 15b to qn 17 I don't understand how to do thanks
Date Posted: 4 years ago
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Not very sure if this is correct, but :
15b
If the first alphabet must be B, we have 2 remaining alphabets to choose from YTES
Number of ways to choose 2 alphabets from the 4 unique alphabets / number of combinations
= 4C2 (4 choose 2)
= 4!/(2!(4-2)!)
= 24/(2! x 2!)
= 24/(2 x 2)
= 24 / 4
= 6
Since the first letter is B and they are to be written in a row,
B_ _
Moment of ways to permute/arrange the 2 alphabets for each combination
= 2!
= 2
Total number of ways = 2 x 6 = 12
15b
If the first alphabet must be B, we have 2 remaining alphabets to choose from YTES
Number of ways to choose 2 alphabets from the 4 unique alphabets / number of combinations
= 4C2 (4 choose 2)
= 4!/(2!(4-2)!)
= 24/(2! x 2!)
= 24/(2 x 2)
= 24 / 4
= 6
Since the first letter is B and they are to be written in a row,
B_ _
Moment of ways to permute/arrange the 2 alphabets for each combination
= 2!
= 2
Total number of ways = 2 x 6 = 12
If you were to list them out :
BYT BTY
BYS BSY
BYE BEY
BTS BST
BTE BET
BES BSE
BYT BTY
BYS BSY
BYE BEY
BTS BST
BTE BET
BES BSE
Thank you. Can you help with qn 17 b and c too :)
17b
S T A T I S T I C S
If all the 'T's must together, we have :
TTT
and the remaining letters are :
S A I S I C S
Number of ways to arrange S A I S I C S
= 7! ÷ 3! ÷ 2!
= 420
_S_ A_ I _S _I _C_ S _
For any permutation of this, we can slot TTT into any of the 8 blank spaces.
So total number of ways
= 420 x 8
= 3360
S T A T I S T I C S
If all the 'T's must together, we have :
TTT
and the remaining letters are :
S A I S I C S
Number of ways to arrange S A I S I C S
= 7! ÷ 3! ÷ 2!
= 420
_S_ A_ I _S _I _C_ S _
For any permutation of this, we can slot TTT into any of the 8 blank spaces.
So total number of ways
= 420 x 8
= 3360
Alternatively, the TTT can be treated as 1 unique unit. So now we have 8 distinct units to permute/permutate.
Number of ways
= 8! ÷ 2! ÷ 3!
= 3360
Number of ways
= 8! ÷ 2! ÷ 3!
= 3360
17c
_S_ A_ I _S _I _C_ S _
If no two 'T's are together, then we slot them individually into one of these blanks.
Number of ways to choose 3 blanks out of 8
= 8C3
= 8! ÷ 3! ÷ (8-3)!
= 8! ÷ 3! ÷ 5!
= 36
Total number of ways = 36 x 420
= 15120
_S_ A_ I _S _I _C_ S _
If no two 'T's are together, then we slot them individually into one of these blanks.
Number of ways to choose 3 blanks out of 8
= 8C3
= 8! ÷ 3! ÷ (8-3)!
= 8! ÷ 3! ÷ 5!
= 36
Total number of ways = 36 x 420
= 15120
Thank you can I ask for 17b right i don't understand the part about slotting TTT into the 8 bank spaces. May I know how these blank spaces are created and may I know the formula u used to compute 7! ÷ 3! ÷ 2! .
For 17c may I know why is it choosing 3 blanks out of 8 instead of 1 thank you!!
For 17c may I know why is it choosing 3 blanks out of 8 instead of 1 thank you!!
17b (secojd method)
You need to keep the TTT together. So we have kind of combined it into 'one big letter' you see.
Or think of it this way :
Treat this TTT as a new letter, eg. X
So now we permutate XSAISICS
That's 8! since there are 8 letters.
But there 3 'S' and 2 'I' . Since there are repeated letters, if you do 8!, We will have some cases that are identical. This is over counting.
So we must divide by the number of permutations for those repeated letters.
You need to keep the TTT together. So we have kind of combined it into 'one big letter' you see.
Or think of it this way :
Treat this TTT as a new letter, eg. X
So now we permutate XSAISICS
That's 8! since there are 8 letters.
But there 3 'S' and 2 'I' . Since there are repeated letters, if you do 8!, We will have some cases that are identical. This is over counting.
So we must divide by the number of permutations for those repeated letters.
If the 3 'S' were distinguishable :
(call them S¹, S² S³)
For a particular permutation,
Eg. S¹TATIS²TICS³
This would be distinct from :
S¹TATIS³TICS²
S²TATIS³TICS¹
S²TATIS¹TICS³
S³TATIS¹TICS²
S³TATIS²TICS¹
These are 6 distinct permutations.
But the issue is, all the 3 'S' are actually indistinguishable. We end up having repeated cases.
So if we just do 8!, we have over counted.
For every other permutation it would also be over counting. We need to divide our number of ways by 3! (=6) since we only want 1 of them.
Similarly,
If the 2 'I' were distinguishable : (call them I¹, I²)
For a particular permutation,
Eg. STATI¹STICS²
This would be distinct from STATI²STI¹CS
These are 2 distinct permutations.
But the issue is, the 2 'I's are actually indistinguishable. We end up having repeated cases.
So if we just do 8!, we have over counted.
For every other permutation it would also be over counting. We need to divide our number of ways by 2! (=2) since we only want 1 of them.
(call them S¹, S² S³)
For a particular permutation,
Eg. S¹TATIS²TICS³
This would be distinct from :
S¹TATIS³TICS²
S²TATIS³TICS¹
S²TATIS¹TICS³
S³TATIS¹TICS²
S³TATIS²TICS¹
These are 6 distinct permutations.
But the issue is, all the 3 'S' are actually indistinguishable. We end up having repeated cases.
So if we just do 8!, we have over counted.
For every other permutation it would also be over counting. We need to divide our number of ways by 3! (=6) since we only want 1 of them.
Similarly,
If the 2 'I' were distinguishable : (call them I¹, I²)
For a particular permutation,
Eg. STATI¹STICS²
This would be distinct from STATI²STI¹CS
These are 2 distinct permutations.
But the issue is, the 2 'I's are actually indistinguishable. We end up having repeated cases.
So if we just do 8!, we have over counted.
For every other permutation it would also be over counting. We need to divide our number of ways by 2! (=2) since we only want 1 of them.
Overall , we need to divide by 2! and by 3! so as to eliminate the repeated cases.
17b (first method on slotting)
SAISICS
This method works by first permutating the 7 remaining letters.
Then, we insert this TTT. We can put it at either end of SAISICS or slot it between any 2 letters.
Example:
STTTISICS
TTTSAISICS
So those blanks are just indicating 8 possible places that you can slot the TTT.
For every of the 420 permutations of SAISICS, there are 8 possible ways to slot.
Therefore we multiply the 420 ways by 8.
An alternative to blanks is to use arrows and place them under the letters but between letters like this :
A B C D
↑ ↑ ↑ ↑ ↑
Each arrow points at each space
17b (first method on slotting)
SAISICS
This method works by first permutating the 7 remaining letters.
Then, we insert this TTT. We can put it at either end of SAISICS or slot it between any 2 letters.
Example:
STTTISICS
TTTSAISICS
So those blanks are just indicating 8 possible places that you can slot the TTT.
For every of the 420 permutations of SAISICS, there are 8 possible ways to slot.
Therefore we multiply the 420 ways by 8.
An alternative to blanks is to use arrows and place them under the letters but between letters like this :
A B C D
↑ ↑ ↑ ↑ ↑
Each arrow points at each space
17c
Why 3 blank spaces instead of 1? Because you have three Ts.
And these three Ts cannot be together.
(Question wording is 'no two Ts are together'.
This means no cases whereby
① two Ts are together, one is separated
② all three Ts together.
Since we need to slot the Ts individually, don't we need to pick 3 spaces for 3 Ts?
Why 3 blank spaces instead of 1? Because you have three Ts.
And these three Ts cannot be together.
(Question wording is 'no two Ts are together'.
This means no cases whereby
① two Ts are together, one is separated
② all three Ts together.
Since we need to slot the Ts individually, don't we need to pick 3 spaces for 3 Ts?
See 1 Answer
Thank you. Can you please help me with 16b too?
This method works as all the letters of BYTES are unique.
If it were BBTES ,BEETY or something like BYTYB you will need to eliminate the repeated cases.
If it were BBTES ,BEETY or something like BYTYB you will need to eliminate the repeated cases.
16b
3 distinct letters, 2 distinct digits, the letters must come first.
To visualise, we can use an example like
ABC12
So we already know the positions of the letters can only be the first 3 characters and the digits can only be at the last 2 characters.
Number of ways to permutate the 3 distinct letters
= 3! = 6
Or
① 3 ways for the 1st letter (A,B,or C)
② For each of the 3 ways, there are 2 ways for the 2nd letter since there are only 2 letters remaining.
If A was first letter, 2nd letter can be B or C
If B was first letter, 2nd letter can be A or C
If C was first letter, 2nd letter can be A or B
③ Only 1 letter is left so 1 way.
Number of ways = 3 x 2 x 1 = 6
(This is basically 3!)
Similarly,
Number of ways to permutate the 2 digits
= 2! = 2
Or
① 2 ways for the 1st digit (1 or 2)
② For either of the 2 ways, there is only 1 digit left. So 1 way for the 2nd digit
Number of ways = 2 x 1 = 2
(This is basically 2!)
Total number of ways
= number of ways for 3 letters x number of ways for 2 digits
= 6 x 2
= 12
3 distinct letters, 2 distinct digits, the letters must come first.
To visualise, we can use an example like
ABC12
So we already know the positions of the letters can only be the first 3 characters and the digits can only be at the last 2 characters.
Number of ways to permutate the 3 distinct letters
= 3! = 6
Or
① 3 ways for the 1st letter (A,B,or C)
② For each of the 3 ways, there are 2 ways for the 2nd letter since there are only 2 letters remaining.
If A was first letter, 2nd letter can be B or C
If B was first letter, 2nd letter can be A or C
If C was first letter, 2nd letter can be A or B
③ Only 1 letter is left so 1 way.
Number of ways = 3 x 2 x 1 = 6
(This is basically 3!)
Similarly,
Number of ways to permutate the 2 digits
= 2! = 2
Or
① 2 ways for the 1st digit (1 or 2)
② For either of the 2 ways, there is only 1 digit left. So 1 way for the 2nd digit
Number of ways = 2 x 1 = 2
(This is basically 2!)
Total number of ways
= number of ways for 3 letters x number of ways for 2 digits
= 6 x 2
= 12
To confirm (for understanding) the full list will be :
ABC12
ACB12
BAC12
BCA12
CAB12
CBA12
ABC21
ACB21
BAC21
BCA21
CAB21
CBA21
ABC12
ACB12
BAC12
BCA12
CAB12
CBA12
ABC21
ACB21
BAC21
BCA21
CAB21
CBA21
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