8 boys and 3 girls are to sit in a row for a photograph. Find the probability that no two girls are together.
Answers
Step-by-step explanation:
We are seeking the situations of the form
(x_1\text{ boys})(\text{a girl})(x_2\text{ boys})(\text{a girl})(x_3\text{ boys})(\text{a girl})(x_4\text{ boys})(x
1
boys)(a girl)(x
2
boys)(a girl)(x
3
boys)(a girl)(x
4
boys),
where x_2,x_3>0.x
2
,x
3
>0. The number of such situations is the number of sequences of nonnegative integers x_1,x_2,x_3,x_4x
1
,x
2
,x
3
,x
4
such that x_1+x_2+x_3+x_4=8x
1
+x
2
+x
3
+x
4
=8 and x_2,x_3>0x
2
,x
3
>0 times the number of permutations between girls and boys, i.e. 8!\cdot 3!8!⋅3!. The number of those sequences is the number of sequences of positive integers y_1,y_2,y_3,y_4y
1
,y
2
,y
3
,y
4
such that y_1+y_2+y_3+y_4=10y
1
+y
2
+y
3
+y
4
=10 with the bijection (x_1,x_2,x_3,x_4)\mapsto(x_1+1,x_2,x_3,x_4+1)=(y_1,y_2,y_3,y_4)(x
1
,x
2
,x
3
,x
4
)↦(x
1
+1,x
2
,x
3
,x
4
+1)=(y
1
,y
2
,y
3
,y
4
), and we indeed have y_1+y_2+y_3+y_4=(x_1+1)+x_2+x_3+(x_4+1)=10y
1
+y
2
+y
3
+y
4
=(x
1
+1)+x
2
+x
3
+(x
4
+1)=10. It's bijective with taking a line of 1010 balls and choosing 33 borders between them with the bijection
(y_1,y_2,y_3,y_4)\mapsto (\text{borders after the balls }y_1,y_1+y_2,y_1+y_2+y_3).(y
1
,y
2
,y
3
,y
4
)↦(borders after the balls y
1
,y
1
+y
2
,y
1
+y
2
+y
3
).
We choose 33 borders from 99 places between balls, which gives \binom93(
3
9
) such sequences. Thus, the probability that no two girls sit near each other equals \b
inom93\cdot 8!\cdot 3!/11!=28/55(
3
9
)⋅8!⋅3!/11!=28/55, where we divide by 11!11! since it's the number of permutations of 8+3=118+3=11 boys and girls.