Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 47,023

PC # 1

There are 7 men and 3 ladies. Find the number of ways in which a committee of 6 persons can be formed if the committee is to have atleast 2 ladies.

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

first, I thank you ganesh for making this new themes.

(But problems and solutions is best, too)

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

So if there are 2 ladies in the com, we have

(7)C(6-2) combinations. Multiply this by the number of 2-ladies:3C2

And simular, if we have 3l:

NUMBER=(7C4)(3C2)+(7C5)(3C3)=21

Is this the answer?

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 47,023

Two ladies must be in the committee.

If two ladies are chosen, the number of combinations is 3C2.

The remaining 4 members can be chosen from the 7 men and the number of combinations is 7C4.

Hence, the number of ways in which two ladies can form part of the committee is 3C2*7C4.

If three ladies are chosen, the number of combinations is 3C3.

The remaining 3 members can be chosen from the remaining 7 men and the number of combinations is 7C3.

Hence, the number of ways in which three ladies can form part of the committee is 3C3*7C3.

The total number of combinations is 3C2*7C4 + 3C3*7C3.

That is, 105 + 35 = 140.

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

One simple muistake may change the answer badly:

(7C4)(3C2)+(7C*5*)(3C3)=21

(7C4)(3C2)+(7C*3*)(3C3)=140

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 47,023

PC # 2

There are 6 books on Economics, 3 on Mathematics and 2 on Accountancy. In how many ways can they be arranged on a shelf if the books of the same subject are always to be together?

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

They must be tigether so we have 3 subjects that must be arranged:

AEM

AME

EAM

EMA

MAE

MEA

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 47,023

Read the question fully......and.....

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 47,023

PC # 3

How many numbers are there between 100 and 1000 such that atleast one of their digits is 6?

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

PC # 3

All of the numbers from 600 to 699 have a 6. This leaves 800 numbers between 100 and 999.

Of those, 1/10 will have 6 as their tens digit. 800/10 = 80, so that leaves 720 more.

Of those, 1/10 will have 6 as their units digit. 720/10 = 72, and adding this to 80 and 100 will give the answer.

100+80+72 = 252.

Why did the vector cross the road?

It wanted to be normal.

Offline

**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 47,023

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 47,023

PC # 4

Bus number plates contain three distinct English alphabets followed by four digits with the first digit not zero. How many different number plates can be formed?

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 47,023

PC # 5

The figures 4, 5, 6,7, and 8 are written in every possible order. How many of the numbers so formed will be greater than 56,000?

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Guess to PC#4

(26) × (25) × (24) × (9) × (10) × (10) × (10)

(Algebra never should have used x's, it sure makes multiplying more confusing.)

**igloo** **myrtilles** **fourmis**

Offline

**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 47,023

You're correct, John! Well done

You could use . instead of x for multiplication sign!

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**landof+****Member**- Registered: 2007-03-24
- Posts: 131

ganesh wrote:

PC # 5

The figures 4, 5, 6,7, and 8 are written in every possible order. How many of the numbers so formed will be greater than 56,000?

First, find ALL the combinations

Since the formula is

where r is

(the number of positions)-1

so now we see: combinations which don't accept 4's at the start and 54's.

and we have:

Done

*Last edited by landof+ (2007-09-16 23:23:23)*

I shall be on leave until I say so...

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Woops I forgot some 57xyz, 58xyz, so add 12 to my answer sorry.

*Last edited by John E. Franklin (2007-09-17 11:11:15)*

**igloo** **myrtilles** **fourmis**

Offline

**landof+****Member**- Registered: 2007-03-24
- Posts: 131

Which is 90, same I suppose

I shall be on leave until I say so...

Offline

**Sudeep****Member**- Registered: 2008-07-21
- Posts: 20

krassi_holmz wrote:

They must be tigether so we have 3 subjects that must be arranged:

AEM

AME

EAM

EMA

MAE

MEA

the answer should be !6 * !3 *!2 * !3

the books can be arranged among themselves as

!6 Economics,

!3 on Mathematics

!2 on Accountancy

and !3 among themselves

Offline

**Sudeep****Member**- Registered: 2008-07-21
- Posts: 20

There are 3 boys and 3 girls. In how many ways can they be arranged so that each boy has at least one girl by his side?

Offline

**Sudeep****Member**- Registered: 2008-07-21
- Posts: 20

There are 10 boxes numbered 1, 2, 3, 10. Each box is to be filled up either with a black or a white ball in such a way that at least 1 box contains a black ball and the boxes containing black balls are consecutively numbered. The total number of ways in which this can be done is..

Offline

**All_Is_Number****Member**- Registered: 2006-07-10
- Posts: 258

ganesh wrote:

Two ladies must be in the committee.

If two ladies are chosen, the number of combinations is 3C2.

The remaining 4 members can be chosen from the 7 men and the number of combinations is 7C4.

Hence, the number of ways in which two ladies can form part of the committee is 3C2*7C4.

If three ladies are chosen, the number of combinations is 3C3.

The remaining 3 members can be chosen from the remaining 7 men and the number of combinations is 7C3.

Hence, the number of ways in which three ladies can form part of the committee is 3C3*7C3.

The total number of combinations is 3C2*7C4 + 3C3*7C3.

That is, 105 + 35 = 140.

There are three ladies from which to choose the two slots designated for ladies. After those two slots are filled, there are eight unchosen members from which to randomly assign the remaining four committee seats. Why isn't the answer nCr(3,2)*nCr(8,4)=3*70=210? What am I missing? If I'm counting 70 possibilities twice, which possibilities am I double counting?

Edit to add: I'm reasonably confident the method I proposed is wrong, and Ganesh's solution is indeed correct. I'm interested in understanding *why* the method I proposed is wrong.

*Last edited by All_Is_Number (2008-10-13 18:32:34)*

*You can shear a sheep many times but skin him only once.*

Offline

**All_Is_Number****Member**- Registered: 2006-07-10
- Posts: 258

Sudeep wrote:

There are 10 boxes numbered 1, 2, 3, 10. Each box is to be filled up either with a black or a white ball in such a way that at least 1 box contains a black ball and the boxes containing black balls are consecutively numbered. The total number of ways in which this can be done is..

*You can shear a sheep many times but skin him only once.*

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

This problem is old but...

The correct answer is 55, by direct count.

You can compute it by:

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

Pages: **1**