Managing "MANAGEMENT"!

Let us count the number of ways of placing the letters M,A,A,M in the 10 possible places, so that no A is next to an M. There are a number of cases, depending on where the M,M go:

• in the 2 cases where M,M are adjacent, and one of them is "terminal" (at the start or end of the word), there are $\binom{7}{2}$ ways of choosing where to put the A,A.
• in the 7 cases where the M,M are adjacent but not terminal, there are $\binom{6}{2}$ ways of choosing where to put the A,A
• in the 2 cases where there is room for just one letter between the M,M, and one of the M is terminal, there are $\binom{6}{2}$ ways of choosing where to put the A,A.
• in the 6 cases where there is room for just one letter between the M,M, and neither M is terminal, there are $\binom{5}{2}$ ways of choosing where to put the A,A.
• in the 1 case where both M are terminal, there are $\binom{6}{2}$ ways of choosing where to put the A,A.
• in the 12 cases where there is room for two or more letters between the M,M, and just one of the M is terminal, there are $\binom{5}{2}$ ways of choosing where to put the A,A.
• in the remaining 15 cases where there is room for two or more letters between the M,M, but neither M is terminal, there are $\binom{4}{2}$ ways of choosing where to put the A,A.

Thus there are $2\binom{7}{2} + 10\binom{6}{2} + 18\binom{5}{2} + 15\binom{4}{2} = 462$ ways of placing the letters M,M,A,A so that no M is next to an A.

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Here is a different argument for obtaining the $462$ ways of arranging the M,M,A,A. This tidies up Kevin's description of my argument. The letters M,M,A,A can occur in one of six orders: AAMM, MMAA, MAAM, AMMA, AMAM, MAMA.

• Suppose that pattern of As and Ms is AAMM. Then we need to put spaces between these letters (with at least one space between the AM in the middle. Suppose the space pattern is (p)A(q)A(r+1)M(s)M(t) (the numbers in brackets represent the number of spaces in that position) where $p,q,r,s,t \ge 0$ and $p+q+r+s+t=5$ . A standard "stars and bars" argument tells us that there are $\binom{9}{4} = 126$ ways of choosing these integers $p,q,r,s,t$ , and so $126$ ways of putting these spaces in. A similar argument holds for the pattern MMAA.
• If the pattern of As and Ms is MAAM, we need a space pattern (p)M(q+1)A(r)A(s+1)M(t), where $p,q,r,s,t \ge 0$ and $p+q+r+s+t=4$ . There are $\binom{8}{4} = 70$ ways of putting these spaces in. A similar argument holds for the pattern AMMA.
• If the pattern of As and Ms is AMAM, we need a space pattern (p)A(q+1)M(r+1)A(s+1)M(t), where $p,q,r,s,t \ge 0$ and $p+q+r+s+t=3$ . Thus there are $\binom{7}{4} = 35$ ways of putting these spaces in. A similar argument holds for the pattern MAMA.

Thus there are, in total $2(126 + 70 + 35) = 462$ ways of placing the As and Ms and allowing spaces for the other letters.

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For each of these $462$ choices, there are $\frac{6!}{2\times2} = 180$ ways of placing the N,N,E,E,G,T in the remaining six places. Thus there are $462\times 180 = 83160$ ways of arranging all ten letters so that no M is next to an A.

Thus there are $\frac{10!}{2\times2\times2\times2} - 83160 = \boxed{143640}$ ways of arranging the letters so that at least one M is next to an A.

The overall idea is to $\\$ All outcomes when no restrictions - All outcomes when no " $m$ " and " $a$ " is adjacent $\\$

Six ways to arrange $mmaa$ which there is always at least one pair of $m$ and $a$ together: $\\$ - $aamm$ , $mmaa$ [One adjacent pair] $\\$ - $maam$ , $amma$ [Two adjacent pair] $\\$ - $mama$ , $amam$ [Three adjacent pair] $\\$

Insert words gap symbols as placeholders to put numbers later. $\\$ - $aa \_\_ mm$ [two 5-letter 1 gap-words] $\\$ - $m \_\_ aa \_\_ m$ [two 6-letter 2 gap-words] $\\$ - $m \_\_ a \_\_ m \_\_ a$ [two 7-letter 3 gap-words] $\\$

Using combinations to calculate the ways to put the rest of vacant spaces into the word as we take out the space reserved for the gap which we will put back to separate $am$ later: $\\$ - $126=9C4$ [All the cases where one fixed space can separate $a$ and $m$ . Ex: ... aa _ mm ...] $\\$ - $70=8C4$ [All the cases where two fixed space can separate $a$ and $m$ . Ex: ... a _ mm _ a ...] $\\$ - $35 = 7C4$ [All the cases where three fixed space can separate​ $a$ and $m$ . Ex: ... a _ m _ a _ m ...] $\\$

Now we only need to fill the letters (eegnnt) into the predetermine vacant spaces: $\\$ $180=\frac {6!}{2! \cdot 2!}$ $\\$

$.\\$ $\frac {10!}{2! \cdot 2! \cdot 2 \cdot 2} = 226800$ $\\$ $226800 - (2 \cdot 126+ 2 \cdot 70+ 2 \cdot 35) \cdot 180 = \boxed{143640}$ $\\$

All Credit goes to $k.stm$ for providing a​ brilliant solution. * k.stm's original solution $\\$ * Thanks to Mark Hennings for kindly pointing​ out the mistakes on my previous version