(a) there are 8C2 = 28 ways of picking 2 girls from 8
And there are 21C4 = 5985 ways of picking 4 boys
Required number of ways for 2g / 4b = 28 * 5985 = 167,580
(b) at least 2 girls means combinations of 2g/4b , 3g,3b , 4g/2b , 5g 1b or
6 girls.
2g/4b = 167,580 ways
3g/3b = 8C3 * 21C3 = 56 * 1330 = 74,480
4g/2b = 8C4* 21C2 = 70 * 210 = 14,700
5g 1b = 8C5* 21 = 56*21 = 1176
6 girls = 8C6 = 28
adding these up we get the answer to (b) which is 257,964
Let q and n represent the number of quarters and nickels respectively.
q=3n and .05n+.25q=1.6 These are the conditions in mathematical terms.
To solve, use the value of q from the first equation in the second equation to get:
.05n+.25(3n)=1.6 carry out indicated multiplication on left side
.05n+.75n=1.6 combine like terms on left side
.8n=1.6 divide both sides by .8
n=2, since q=3n
q=6
So Peggy had 2 nickels and 6 quarters.
Factor it out first so:
y=(x+13)(x-2)
Then you know that x= -13 and 2 to make the equation 0
Hope this helps!
