combination for letters with
$begingroup$
Think about all the meaningful and meaningless 9-letter words that can be obtained by using all the letters in ALAFRANGA. Some examples are AAFARNLAG, RANALAGFA, NAAALAGFR etc. Of all these words, in how many of them two or more A’s are not next to each other? For example, RANALAGFA is OK, AAFARNLAG and NAAALAGFR are not OK.
combinatorics
edited Jan 6 at 17:17
Digitalis
529216
asked Jan 6 at 16:45
Ferda TaşFerda Taş
42
$endgroup$
add a comment |
$begingroup$
Think about all the meaningful and meaningless 9-letter words that can be obtained by using all the letters in ALAFRANGA. Some examples are AAFARNLAG, RANALAGFA, NAAALAGFR etc. Of all these words, in how many of them two or more A’s are not next to each other? For example, RANALAGFA is OK, AAFARNLAG and NAAALAGFR are not OK.
combinatorics
edited Jan 6 at 17:17
Digitalis
529216
asked Jan 6 at 16:45
Ferda TaşFerda Taş
42
$endgroup$
add a comment |
$begingroup$
Think about all the meaningful and meaningless 9-letter words that can be obtained by using all the letters in ALAFRANGA. Some examples are AAFARNLAG, RANALAGFA, NAAALAGFR etc. Of all these words, in how many of them two or more A’s are not next to each other? For example, RANALAGFA is OK, AAFARNLAG and NAAALAGFR are not OK.
combinatorics
edited Jan 6 at 17:17
Digitalis
529216
asked Jan 6 at 16:45
Ferda TaşFerda Taş
42
$endgroup$
Think about all the meaningful and meaningless 9-letter words that can be obtained by using all the letters in ALAFRANGA. Some examples are AAFARNLAG, RANALAGFA, NAAALAGFR etc. Of all these words, in how many of them two or more A’s are not next to each other? For example, RANALAGFA is OK, AAFARNLAG and NAAALAGFR are not OK.
combinatorics
combinatorics
edited Jan 6 at 17:17
Digitalis
529216
asked Jan 6 at 16:45
Ferda TaşFerda Taş
42
edited Jan 6 at 17:17
Digitalis
529216
asked Jan 6 at 16:45
Ferda TaşFerda Taş
42
edited Jan 6 at 17:17
Digitalis
529216
edited Jan 6 at 17:17
Digitalis
529216
edited Jan 6 at 17:17
Digitalis
529216
529216
asked Jan 6 at 16:45
Ferda TaşFerda Taş
42
asked Jan 6 at 16:45
Ferda TaşFerda Taş
42
asked Jan 6 at 16:45
Ferda TaşFerda Taş
42
42
add a comment |
add a comment |
1 Answer
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$begingroup$
You can use the Stars and Bars technique. In this case, the "bars" are the $A$s, and they divide the other letters. Notice that the other letters are ${L, F, R, N, G}$ (all distinct). We can place bars anywhere in the spacing outside the edge and between letters, like:
_*_*_*_*_*_
Notice that there are $6$ slots, from which we must select $4.$ This can be done in $begin{pmatrix} 6 \ 4 end{pmatrix} = 15$ ways. The other letters (represented by asterisks) can be filled in $5! = 120$ ways. So the total number of such strings is $15 cdot 120 = boxed{1800}.$
answered Jan 6 at 17:06
K. JiangK. Jiang
3,0311513
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add a comment |
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$begingroup$
You can use the Stars and Bars technique. In this case, the "bars" are the $A$s, and they divide the other letters. Notice that the other letters are ${L, F, R, N, G}$ (all distinct). We can place bars anywhere in the spacing outside the edge and between letters, like:
_*_*_*_*_*_
Notice that there are $6$ slots, from which we must select $4.$ This can be done in $begin{pmatrix} 6 \ 4 end{pmatrix} = 15$ ways. The other letters (represented by asterisks) can be filled in $5! = 120$ ways. So the total number of such strings is $15 cdot 120 = boxed{1800}.$
answered Jan 6 at 17:06
K. JiangK. Jiang
3,0311513
$endgroup$
add a comment |
$begingroup$
You can use the Stars and Bars technique. In this case, the "bars" are the $A$s, and they divide the other letters. Notice that the other letters are ${L, F, R, N, G}$ (all distinct). We can place bars anywhere in the spacing outside the edge and between letters, like:
_*_*_*_*_*_
Notice that there are $6$ slots, from which we must select $4.$ This can be done in $begin{pmatrix} 6 \ 4 end{pmatrix} = 15$ ways. The other letters (represented by asterisks) can be filled in $5! = 120$ ways. So the total number of such strings is $15 cdot 120 = boxed{1800}.$
answered Jan 6 at 17:06
K. JiangK. Jiang
3,0311513
$endgroup$
add a comment |
$begingroup$
You can use the Stars and Bars technique. In this case, the "bars" are the $A$s, and they divide the other letters. Notice that the other letters are ${L, F, R, N, G}$ (all distinct). We can place bars anywhere in the spacing outside the edge and between letters, like:
_*_*_*_*_*_
Notice that there are $6$ slots, from which we must select $4.$ This can be done in $begin{pmatrix} 6 \ 4 end{pmatrix} = 15$ ways. The other letters (represented by asterisks) can be filled in $5! = 120$ ways. So the total number of such strings is $15 cdot 120 = boxed{1800}.$
answered Jan 6 at 17:06
K. JiangK. Jiang
3,0311513
$endgroup$
You can use the Stars and Bars technique. In this case, the "bars" are the $A$s, and they divide the other letters. Notice that the other letters are ${L, F, R, N, G}$ (all distinct). We can place bars anywhere in the spacing outside the edge and between letters, like:
_*_*_*_*_*_
Notice that there are $6$ slots, from which we must select $4.$ This can be done in $begin{pmatrix} 6 \ 4 end{pmatrix} = 15$ ways. The other letters (represented by asterisks) can be filled in $5! = 120$ ways. So the total number of such strings is $15 cdot 120 = boxed{1800}.$
answered Jan 6 at 17:06
K. JiangK. Jiang
3,0311513
answered Jan 6 at 17:06
K. JiangK. Jiang
3,0311513
answered Jan 6 at 17:06
K. JiangK. Jiang
3,0311513
answered Jan 6 at 17:06
K. JiangK. Jiang
3,0311513
3,0311513
add a comment |
add a comment |
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