Permutations And Combinations - Class 11 Commerce Maths - Extra Questions

In a polygon, no three diagonals are concurrent. If the total number of points of intersection of diagonals interior to the polygon is $$70$$ then the number of diagonals of the polygon is

The number of intersecting diagonals can be found by $${^nC_4}$$
or $$n(n-1)(n-2)(n-3)=70\times 24=1680$$
$$n=8$$
Now number of diagonals $$={^8C_2}-8$$
                                                             $$\downarrow$$
                                                  $$8$$ will be collinear
$$\Rightarrow 28-8=20.$$
Hence, the answer is $$20$$



Find the last $$3$$ digits of the numbers $$\displaystyle { 2003 }^{ { 2002 }^{ 2001 } }$$

Given 
$$2003^{2002^{2001}}$$

$$2003\equiv 3 mod 1000$$
SO
$$3^{(2002^{2001})} mod 1000$$. 

$$\lambda (1000) = 1000 \left(1 - \dfrac{1}{2}\right) \left(1 - \dfrac{1}{5}\right) = 400$$. 

So $$3^{400} ≡ 1 mod 1000$$. 

So we need to find 

$$2002^{2001} mod 400$$. 

$$2002 ≡ 2 mod 400$$, so we find 

$$2^{2001} mod 400$$. 

$$\lambda(400) = 400\left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{5}\right) = 160$$. 

So $$2^{160} ≡ 1 mod 400$$. 

$$2001 = 160 \times 12 + 81$$ so $$2^{2001} mod 400$$ is congruent to 

$$2^{81} mod 400$$. 

$$400 = 25 \times 16$$, and 

$$2^{81} mod 16 ≡ 0$$ (as $$2^4 = 16$$)
 
$$2^{81} mod 25 ≡ 2 mod 25$$ (as $$\lambda(25) = 20$$). 

The only number less than 400 that is 2 more than a multiple of 5, and divisible by 16 is 352. Therefore, 

$$2^{81} mod 400 = 352$$. 

So 

$$3^{(2002^{2001})} ≡ 3^{352} mod 1000$$. 

$$1000 = 125 \times 8$$ 


$$\lambda(8) = 4 \rightarrow 3^4 ≡ 1 mod 8 $$
$$\Rightarrow 3^{352} ≡ 3^{(4\times 88)} ≡ 1 mod 8$$. 

$$\lambda(125) = 100 \rightarrow 3^{100} ≡ 1 mod 125$$ 
$$\Rightarrow 3^{352} mod 125 ≡ 3^{52} mod 125$$. 

$$3^{52} mod 125$$ 
$$≡ 9^{26} mod 125$$ 
$$≡ 81^{13} mod 125$$ 

$$81^2 ≡ 6561 ≡ 61 mod 125$$ 
$$61^² ≡ 3721 ≡ 96 mod 125$$ 

So 

$$81^{13} mod 125$$ 
$$≡ 81 \times 81^{12} mod 125 $$
$$≡ 81 \times (81^4)^3 mod 125$$ 
$$≡ 81 \times 96^3 mod 125$$ 

$$96^² ≡ 9216 ≡ 91 mod 125$$ 
and 
$$81\times 96 ≡ 7776 ≡ 26 mod 125$$ 

so 

$$81 \times 96^3 mod 125$$ 
$$≡ 26 \times 91 mod 125$$ 
$$≡ 2366 mod 125 $$
$$≡ 116 mod 125. $$

The only number simultaneously of the form 8k + 1 and 125m + 116 that is also less than 1000 is 241. Thus 

$$3^{352} mod 1000 ≡ 241$$ 
Hence answer is $$241$$



Find the value of
If $$^{15}C_{3r} : ^{15}C_{r + 1} = 11:3$$ find $$r$$.

Given:
$$\dfrac{{}^{15}C_{3r}}{{}^{15}C_{r+1}}=\dfrac{\dfrac{(15)!}{(3r)!(15-3r)!}}{\dfrac{(15)!}{(r+1)!(15-r-1)!}}=\dfrac{11}{3}$$
We get
$$\dfrac{(r+1)!(15-r-1)!}{(3r)!(15-3r)!}=\dfrac{11}{3} $$
When $$r=1$$
$$\dfrac{(2)!(15-1-1)!}{(3)!(15-3)!}=\dfrac{13}{3} $$
When $$r=2$$
$$\dfrac{(3)!(12)!}{(6)!(9)!}={11}$$
When $$r=3$$
$$\dfrac{(4)!(11)!}{(9)!(6)!}=\dfrac{11}{3}$$
So $$r=3$$ is the answer.


Class 11 Commerce Maths Extra Questions