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## Here's the Solution to this Question

1.Number of ways in which he can do 2 asignments $= ^5P_2=\dfrac{5!}{(5-2)!}=\dfrac{120}{6}=20$

2.Number of ways $= ^{10}C_5\times ^{6}C_3=252\times 20=5040$

3.No. of ways of selcting 3 balls

a)if they are of different colour $=^6C_1\times ^5C_1 \times ^3C_1=90$

b)If they all are red $= ^6C_3=20$

c)if two are blues and one is white $= ^5C_2\times ^3C_1=10\times 3=30$

d)if exactly 2 are blue $= ^5C_2\times ^9C_1=10\times 9=90$

e)if none is white $= ^{11}C_3=132$

f)if at least two are whute $=^3C_2\times ^{11}C_1=3\times11=33$

Binomial coeffecients'

1. $(2m-2b)^3$

$=^3C_0(2m)^0(2b)^3+^3C_1(2m)^1(2b)^2+^3C_2(2m)^2(2b)^1+^3C_3(2m)^3(2b)^0\\ =8b^3+24mb^2+24m^2b+8m^3$

2. Given expression is -

$(a-10b)^{10}$

The general term is-

$T_{r+1}=^{10}C_r(a)r(-10b)^{10-r}$

at r=5, coefficient of $a^5b^5= ^{10}C_5\times (-10)^5=-25200000$

3. The given expression is-

$(3x-4y)^{15}$

The $5^{th}$ term is = $^{15}C_4 (3x)^4 (-4y)^{15-4}$

Peigenhole principle

In this case, there are 26 pigeonholes (one for each letter in the English alphabet) but 27 pigeons, so we can immediately see that there must be at least one bucket with more than one word in it. But a bucket with more than one word implies that there are at least two words that start with the same letter.