# Thread: Refreshment of permutation and combination

1. ## Refreshment of permutation and combination

(1) How many four letter words can be formed using the letters of the word EQUATION so that the letter Q is included in each word ?

isn' t the answer 7x6x5 = 210 ... Am i correct ?

(2) How many five letter words can be formed by picking 3 vowels and 2 consonants from the word REACTION such that two consonants always remain together ?

(1) How many four letter words can be formed using the letters of the word EQUATION so that the letter Q is included in each word ?

isn' t the answer 7x6x5 = 210 ... Am i correct ?
No

If Q is at the first place then you have 7 choices for the second letter then 6 choices for the third one and finally 5 choices for the fourth one
That makes 7x6x5 = 210
But if Q is at the second place you have also 210 possibilities. The same if Q is at the third place and the same for the fourth place.
Therefore there are 210 x 4 = 840 words

Another way is to calculate the number of group of 4 letters containing Q
There are $\binom{7}{3} = \frac{7!}{3!\: 4!} = 35$ groups of 4 letters with Q (the same number as the number of groups of 3 letters among 7 letters)
For each group of 4 letters you can build 4! = 24 words
The total number of words is 35 x 24 = 840

Another approach . . .

(1) How many 4-letter words can be formed using the letters of the word EQUATION
so that the letter Q is included in each word?
The letter Q will be included.
Select 3 letters from the other 7 letters: . ${7\choose 3} \,=\,{\color{blue}35}$ ways.

We have a set of 4 letters; they can be arranged in: $4! \,=\,{\color{blue}24}$ ways.

Therefore, there are: . $35 \times 24 \:=\:\boxed{{\color{blue}840}}$ possible words.

(2) How many 5-letter words can be formed by picking 3 vowels and 2 consonants
from the word REACTION such that two consonants always remain together?
There are 4 vowels and 4 consonants.

Select 3 of the 4 vowels: . ${4\choose3} \,=\,4$ ways.
Select 2 of the 4 consonants: . ${4\choose2} \,=\,6$ ways.

. . We have: . $4 \times 6 \:=\:{\color{blue}24}$ selections of letters.

We have 3 vowels and 2 consonants: . $\{V,V,V,C,C\}$

Duct-tape the two consonants together: . $\left\{V,V,V,\boxed{CC}\right\}$

These four "letters" can be arranged in: $4! \,=\,24$ ways.
But the two constants could be ordered either $C_1C_2$ or $C_2C_1$.
. . Hence, there are: . $2 \times 24 \:=\:{\color{blue}48}$ orders.

Therefore, there are: . $24 \times 48 \:=\:\boxed{{\color{blue}1152}}$ possible words.

4. Originally Posted by Soroban

Another approach . . .

The letter Q will be included.
Select 3 letters from the other 7 letters: . ${7\choose 3} \,=\,{\color{blue}35}$ ways.

We have a set of 4 letters; they can be arranged in: $4! \,=\,{\color{blue}24}$ ways.

Therefore, there are: . $35 \times 24 \:=\:\boxed{{\color{blue}840}}$ possible words.

Not so new !
See my post