Class 11 Commerce Maths - Permutations And Combinations

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

${^nC_4}$

$n(n-1)(n-2)(n-3)=70\times 24=1680$

$n=8$

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

$2003^{2002^{2001}}$

$2003\equiv 3 mod 1000$

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

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

$3^{400} ≡ 1 mod 1000$ .

So we need to find

$2002^{2001} mod 400$ .

$2002 ≡ 2 mod 400$ , so we find

$2^{2001} mod 400$ .

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

$2^{160} ≡ 1 mod 400$ .

$2001 = 160 \times 12 + 81$ 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,

$2^{81} mod 400 = 352$ .

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

$1000 = 125 \times 8$

$\lambda(8) = 4 \rightarrow 3^4 ≡ 1 mod 8$

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

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

$\Rightarrow 3^{352} mod 125 ≡ 3^{52} mod 125$ .

$3^{52} mod 125$

$≡ 9^{26} mod 125$

$≡ 81^{13} mod 125$

$81^2 ≡ 6561 ≡ 61 mod 125$

$61^² ≡ 3721 ≡ 96 mod 125$

$81^{13} mod 125$

$≡ 81 \times 81^{12} mod 125$

$≡ 81 \times (81^4)^3 mod 125$

$≡ 81 \times 96^3 mod 125$

$96^² ≡ 9216 ≡ 91 mod 125$

and

$81\times 96 ≡ 7776 ≡ 26 mod 125$

$81 \times 96^3 mod 125$

$≡ 26 \times 91 mod 125$

$≡ 2366 mod 125$

$≡ 116 mod 125.$

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.

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

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

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

Class 11 Commerce Maths Extra Questions

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 7070 then the number of diagonals of the polygon is

Find the last 33 digits of the numbers 200320022001\displaystyle { 2003 }^{ { 2002 }^{ 2001 } }

Find the value ofIf 15C3r:15Cr+1=11:3^{15}C_{3r} : ^{15}C_{r + 1} = 11:3 find rr.

Class 11 Commerce Maths Extra Questions

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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

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

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