Permutations and Combinations Test 26

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Permutations and Combinations Test 26

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Weekly Quiz Competition

Question 1
1 / -0

Robert was asked to made a $$5$$ digit number from the digits $$2$$ and $$4$$ such that first digit cannot be $$4$$. Find the number of such $$5$$ digit numbers that can be formed .

A
$$10$$

B
$$12$$

C
$$16$$

D
$$18$$

E
None of these

Solution

Total $$5$$ digit numbers $$=1\times 2\times 2\times 2\times 2$$

$$=2^{4}$$

$$=16$$

$$C$$ is correct
Question 2
1 / -0

If A and B are the sums of odd and even terms respectively in the expansion of $$(x+a)^n$$,then $$(x+a)^{2n} -(x-a)^{2n}$$ is equal to:

A
$$4(A+B)$$

B
$$4(A-B)$$

C
$$AB$$

D
$$4AB$$

Solution

$$A$$=sum of odd terms
Let n is odd,let total $$T_{n+1}$$ terms
$$A=T_1+T_3+...T_n$$
$$A=^nC_0+^nC_2+....+^nC_{n-1}$$
$$A=\dfrac{2^n}{2}=2^{n-1}$$
similarly $$B=T_2+T_4+...T_{n+1}$$
$$B=^nC_1+^nC_3+....+^nC_n$$
$$B=\dfrac{2^n}{2}=2^{n-1}$$
$$A+B=2^n$$
$$(x+a)^{2n} -(x-a)^2n \Rightarrow (x+a+x-a)^n (x+a-x+a)^n$$
$$(2x)^n (2a)^n \Rightarrow 2^{2n} x^n a^n=4^n a^n x^n$$
Considering this term being $$4^n$$
$$AB=\dfrac{2^{2n}}{4}$$
$$4AB=2^{2n}=4^n$$
Therefore,$$(x+a)^{2n}-(x-a)^{2n}=4AB$$
Question 3
1 / -0

PQRS is a quadrilateral having $$3, 4, 5, 6$$ points in PQ, QR, RS and SP respectively. The number of triangles with vertices on different sides is?

A
$$220$$

B
$$270$$

C
$$282$$

D
$$342$$

Solution

The number of triangles with vertices on side, $$AB,BC,CD=3{C}_{1}\times 4C_1\times 5C_1$$
$$\therefore$$ Total no. of triangles $$=3C_1\times 4C_1\times 5C_1+3C_1\times 4C_1\times 6C_1+3C_1\times 5C_1\times 6C_1+ 4C_1\times 5C_1\times 6C_1\\=3\times 4\times 5+3 \times 4\times 6+3 \times 5\times 6+4\times 5\times 6\\=342$$
Question 4
1 / -0

How many $$3$$-digit even numbers can be formed from the digits $$1, 2, 3, 4, 5, 6$$ if the digits can be repeated?

A
108

B
98

C
72

D
112

Solution

Only $$3$$ numbers are possible at units place $$(2,4,6)$$ as we need even numbers. But at $$10's$$ place and $$100's$$ place all $$6$$ are possible.
No. of digit even no's$$=3\times 6\times 6=108$$
Question 5
1 / -0

The value of $$\sum\limits_{r = 1}^{10} r .\frac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}}$$ is equal to

A
$$5(2n-9)$$

B
$$10n$$

C
$$9(n-4)$$

D
$$n-2$$

Solution

CORRECTION: $$10n-45 = 5(2n-9)$$ Option A correct
Question 6
1 / -0

The number of $$5$$ digit telephone numbers having least one of their digits repeated is

A
$$ 90,000$$

B
$$ 100,000$$

C
$$ 30,240$$

D
$$ 69,760$$

Solution

Consider the given question,
Numbers with repetition$$=10\times 10\times 10\times 10\times 10$$
$$=1,00,000$$

Numbers without repetition $$=10\times 9\times 8\times 7\times 6$$
$$=30240$$

Number with one digit is repeated$$=100000-30240$$
$$=69760$$

Hence, the number of 5 digit telephone numbers having least one of their digits repeated is $$69760$$
Question 7
1 / -0

Number of ways in which $$7$$ green bottles and $$8$$ blue bottles can be arranged in a row if exactly $$1$$ pair of green bottles is side by side is (Assume all bottles to be a like except for the colour).

A
$$84$$

B
$$360$$

C
$$504$$

D
None of the above

Solution

Consider the two green bottles as one entity. Let's call it $$GG$$.

We will also refer to the green bottle as: $$G$$, and to the blue bottle as: $$B$$.

So now we have $$14$$ units to arrange ( 1 green pair bottle + the remaining 5 green bottles + the 8 blue bottles):

$$GG, G, G, G, G, G, B, B, B, B, B, B, B, B$$

Now place the 8 blue bottles in a row, such that between each bottle has a gap before and after it:

_B_B_B_B_B_B_B_B_

Now we need to take the six green units (the 1 pair and 5 bottles) and place them in six of the nine gaps.

Therefore, the solution is: $$^9C_6 = 84 ways$$.
Question 8
1 / -0

There are $$10$$ points in a plane of which no $$3$$ points are collinear and $$4$$ points are concylic. No. of different circles that can be drawn through atleast $$3$$ points of these given points is

A
$$\left( ^{ 8 }{ C }_{ 3 }-^{ 6 }{ C }_{ 3 } \right) +1$$

B
$$\left( ^{ 10 }{ C }_{ 3 }-^{ 4 }{ C }_{ 3 } \right) +1$$

C
$$\left( ^{ 6 }{ C }_{ 3 }-^{ 4 }{ C }_{ 3 } \right) +1$$

D
$$\left( ^{ 4 }{ C }_{ 3 }-^{ 2 }{ C }_{ 3 } \right) +1$$

Solution

We need at least $$3$$ points to form a circle:$$^{ 10 }{ C }_{ 3 }$$
Out of these $$4$$ points are already concyclic so we must remove $$^{ 4 }{ C }_{ 3 }-1$$ cases.
$$^{ 10 }{ C }_{ 3 }-\left( ^{ 4 }{ C }_{ 3 }-1 \right) $$
Answer is:$$\left( ^{ 10 }{ C }_{ 3 }-^{ 4 }{ C }_{ 3 } \right) +1$$
Question 9
1 / -0

The sides AB, BC, CA of a triangle ABC have $$3,4$$ and $$5$$ interior points respectively on them. The number of triangles that can be constructed using these points as vertices is-

A
$$205$$

B
$$210$$

C
$$315$$

D
$$216$$

Solution

$$\rightarrow $$Suppose we had $$12$$ points, none of them collinear, number of possible triangles $$=^{ 12 }{ C }_{ 3 }$$
$$\rightarrow $$If out of these $$3,4$$ and $$5$$ points were chosen separately to lie on each side of a triangle ABC, then number of possible triangles
$$=^{ 12 }{ C }_{ 3 }-^{ 3 }{ C }_{ 3 }-^{ 5 }{ C }_{ 3 }-^{ 4 }{ C }_{ 3 }$$
$$=220-1-10-4$$
$$=205$$
Question 10
1 / -0

$$\frac{{({C_0} + {C_1})({C_1} + {C_2})({C_2} + {C_3}).....({C_{n - 1}} + {C_n})}}{{{C_0} + {C_1}({C_2}......{C_n})}} = $$

A
$$\frac{{{{(n + 1)}^n}}}{{n!}}$$

B
$$\frac{{{{(n + 2)}^n}}}{{n!}}$$

C
$$\frac{{{{(n + 1)}^{n - 1}}}}{{n!}}$$

D
$$\frac{{{{(n - 1)}^n}}}{{n!}}$$

Solution

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Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 26

Result

Permutations and Combinations Test 26

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SHARING IS CARING

Question 1

$$10$$

$$12$$

$$16$$

$$18$$

None of these

Solution

Question 2

$$4(A+B)$$

$$4(A-B)$$

$$AB$$

$$4AB$$

Solution

Question 3

$$220$$

$$270$$

$$282$$

$$342$$

Solution

Question 4

108

98

72

112

Solution

Question 5

$$5(2n-9)$$

$$10n$$

$$9(n-4)$$

$$n-2$$

Solution

Question 6

$$ 90,000$$

$$ 100,000$$

$$ 30,240$$

$$ 69,760$$

Solution

Question 7

$$84$$

$$360$$

$$504$$

None of the above

Solution

Question 8

$$\left( ^{ 8 }{ C }_{ 3 }-^{ 6 }{ C }_{ 3 } \right) +1$$

$$\left( ^{ 10 }{ C }_{ 3 }-^{ 4 }{ C }_{ 3 } \right) +1$$

$$\left( ^{ 6 }{ C }_{ 3 }-^{ 4 }{ C }_{ 3 } \right) +1$$

$$\left( ^{ 4 }{ C }_{ 3 }-^{ 2 }{ C }_{ 3 } \right) +1$$

Solution

Question 9

$$205$$

$$210$$

$$315$$

$$216$$

Solution

Question 10

$$\frac{{{{(n + 1)}^n}}}{{n!}}$$

$$\frac{{{{(n + 2)}^n}}}{{n!}}$$

$$\frac{{{{(n + 1)}^{n - 1}}}}{{n!}}$$

$$\frac{{{{(n - 1)}^n}}}{{n!}}$$

Solution

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