# “ALGEBRA” so that no two vowels come together?

_L_G_B_R_

Possible ways to arrange consonants among themselves $= 4!$ Now we need to find the number of ways we can arrange $3$ vowels in $5$ places. This could be done by ${5\choose3}\frac{3!}{2!}$

Hence the total number of ways = $$4! * {5\choose3} * \frac{3!}{2!} = 24 *20 * 3 = 1440$$ ways.

However, when I tried doing it using inclusion and exclusion, I don't get the same answer. Can someone correct me, please??

$X$ - Represents the set of arrangements for the word "$ALGEBRA$" = $\frac{7!}{2!}$

$A$ - Set of arrangements where $2A$ are together = $6!$ ( We take AA as one unit)

$B$ - Set of arrangements where $A$ and $E$ are together $6!*2!$ (AE/EA)

$C$ - Set of arrangements where $2A$ and $E$ are together $5!*3!$ (AAE, EAA, AEA)

$$|A \cap B| = |A \cap C|= |B \cap C|= |A \cap B \cap C| = 5! * 3!$$

$$|A \cup B \cup C|=|A|+|B|+|C| -|A\cap B|-|A \cap C|-|B \cap C|+ |A \cap B \cap C|$$

Answer $$2440 - ( 6! + 6!*2! + 5!*3! - 3(5!*3!) + 5!*3! = 2440-1440 = 1000...$$

EDIT: So other people could know what I did wrong, I will only add the changes that will make it work without editing the real question... I had a small mistake in method 1 miscalculating ${5\choose3}$ as 20 not 10 so the real answer should be 720 not 1440.

Method 2 arrangements for AAE EAA AEA should be $3$ and not $3!$

Thanks to @paw88789 and @Niing

• You said $3!$ and wrote $(AAE, EAA, AEA)$, I think that's the little error. – linear_combinatori_probabi May 3 '18 at 12:31
• Note that ${5 \choose 3} = 10$, not $20$. So your first calculation should work out to $720$, not $1440$. – paw88789 May 3 '18 at 12:35
• @paw88789: Combine with yours that's why wolfram tells me the answer is $720$. – linear_combinatori_probabi May 3 '18 at 12:36
• lol that was the problem. After changing 3! to 3, it gives 720. I also fixed it from the top. Thanks so much guys!!!! – nano mano May 3 '18 at 13:07

Method 1: We arrange the consonants, then place the vowels in the spaces between and at the ends of the row.

There are $4!$ ways of arranging the four distinct consonants B, G, L, R. For each such arrangement, there are five spaces in which we could place the vowels, three between successive consonants and two at the ends of the row. To separate the vowels, we must place the three vowels in three of these five spaces. Choose two of the five spaces for the As and one of the remaining three spaces for the E. Thus, the number of admissible arrangements is $$4!\binom{5}{2}\binom{3}{1} = 720$$

Method 2: We use the Inclusion-Exclusion Principle.

There are seven letters in total, including two As, one B, one E, one G, one L, and one R. There are $\binom{7}{2}$ ways to choose two of the seven positions in the arrangement for the As. The remaining five letters can be placed in the remaining five positions in $5!$ ways. Hence, there are a total of $$\binom{7}{2}5! = \frac{7!}{2!5!} \cdot 5! = \frac{7!}{2!}$$ ways of arranging the letters of the word ALGEBRA.

From these, we must subtract those arrangements in which one or more pairs of vowels are consecutive.

A pair of vowels is consecutive: Two As are consecutive or an A and E are consecutive.

Two As are consecutive: We have six objects to arrange. They are AA, B, E, G, L, R. Since they are distinct, they can be arranged in $6!$ ways.

An A and an E are consecutive: We have six objects to arrange. They are B, E, G, L, R, and a block containing an A and an E. The six objects are distinct, so they can be arranged in $6!$ ways. The A and E can be arranged within the block in $2!$ ways. Hence, there are $6!2!$ such arrangements.

If we subtract the number of arrangements in which a pair of vowels is consecutive from the total, we will have subtracted each arrangement in which two pairs of vowels are consecutive twice, once for each way we could have designated one of those pairs as the pair of consecutive vowels. Since we only want to subtract such arrangements once, we must add them to the total.

Two pairs of consecutive vowels: Since there are only three vowels, this can only occur if the three vowels are consecutive. We have five objects to arrange, B, G, L, R, and the block of three vowels. Since the objects are distinct, they can be arranged in $5!$ ways. The three vowels can be arranged in three ways: AAE, AEA, EAA. Hence, the number of such arrangements is $5! \cdot 3$.

By the Inclusion-Exclusion Principle, the number of admissible arrangements is $$\frac{7!}{2!} - 6! - 6!2! + 5! \cdot 3 = 720$$

• Thanks for further explanation :) – nano mano May 3 '18 at 13:10