Permutations Help :D (1 Viewer)

[mawissah]

I'm not so good with permutations and combinations and i need help

a) How many seven letter words can be formed without repetition from the letters of the word INCLUDE?
b) How many of these do not begin with I?
c) How many end with L?
d) How many have the vowels and consonants alternating?
e) How many have the C immediately following the D?
f) How many have the letters N and D separated by exactly two letters?
g) How many have the letters N and D separated by more than two letters?

 

[Carrotsticks]

a) 7! (since 7 distinct letters)

b) 7! - 6! (subtract letters beginning with I from the total)

c) 6! (fix L at the end so we have 6 letters remaining)

d) 3!4! (only one case of alternating, so we arrange 3 vowels then 4 consonants)

e) 6! (group the C and D, so we have 6 groups in total including the pair.)

f) 4! x 5C2 x 2! x 2! (group the N and D with the two in between. We now have 4 groups in total, including the massive one. Permute them. Then 5C2 two people out of 5 possible people to go in between, then swap the 2 people, then swap N and D)

g) 7! - (part (e) + part (f)) (complementary event of (e) and (f) combined)

EDIT: Added reasons.

EDIT2: Added corrections.

 

[mawissah]

thankyou!!!!! ^^

 

[braintic]

a) 7! (since 7 distinct letters)

b) 7! - 6! (subtract letters beginning with I from the total)

c) 6! (fix L at the end so we have 6 letters remaining)

d) 3!4! (only one case of alternating, so we arrange 3 vowels then 4 consonants)

e) 6!2! (group the C and D, so we have 6 groups in total including the pair. Arrange groups, then swap the C and D)

f) 4! x 5C2 x 2! (group the N and D with the two in between. We now have 4 groups in total, including the massive one. Permute them. Then 5C2 two people out of 5 possible people to go in between, then swap the 2 people)

g) 7! - (part (e) + part (f)) (complementary event of (e) and (f) combined)

EDIT: Added reasons.

I'm afraid e, f, and g are not right.

(e) C had to be immediately FOLLOWING D, so we need to lose the 2!

(f) This time, no order is given for N and D. So we need to multiply by 2 TWICE - once for the N and D, and once for the pair inside.
Another way: The N and D can fill 4 positions: 1&4, 2&5, 3&6 or 4&7. Then the N and D can be swapped over. Then arrange the remaining 5. So 4 x 2 x 5!.
Either way, the answer is twice yours.

(g) Due to the the error in (e), it should be 7! - (2 x part (e) + part (f))

 

[Carrotsticks]

Thanks for pointing out the errors.

Sillies are the bane of my life =)

 

[HoChiMinh]

I'm afraid e, f, and g are not right.

(e) C had to be immediately FOLLOWING D, so we need to lose the 2!

(f) This time, no order is given for N and D. So we need to multiply by 2 TWICE - once for the N and D, and once for the pair inside.
Another way: The N and D can fill 4 positions: 1&4, 2&5, 3&6 or 4&7. Then the N and D can be swapped over. Then arrange the remaining 5. So 4 x 2 x 5!.
Either way, the answer is twice yours.

(g) Due to the the error in (e), it should be 7! - (2 x part (e) + part (f))

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In part (f), the N and D can also take positions 2&5, 2&6, 2&7, 3&6, 3&7 and 4&7. So there are a total of 10 positions for the N and D instead of 4. Am I correct?