A lady has fine gloves and hats in her closet- 14 blue, 20 red, and 18 yellow. The lights are out and it is totally dark. In spite of the darkness, she can make out the difference between a hat and a glove. She takes out an item out of the closet only if she is sure that if it is a glove. How many gloves must she take out to make sure she has a pair of each color?
In this question we are not asked how many ways we can select 3 numbers out of 8.
But how many different numbers can be expressed as a sum of three numbers from the given set.
For example, 3 + 10 + 31 = 3 + 17 + 24 = 47. So 47 can be expressed as a sum of 3 numbers in two different ways but 47 should be considered as only one number.
Now the minimum number that can be expressed as a sum of 3 numbers = 30. (3+10+17)
The next number is 37.
Similarly the largest number is 38 + 45 + 52 = 135.
So there exists many numbers in between, with common difference of 7.
Total numbers = (l?a)/d +1 = (135-30)/7 +1= 15+1=16.
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A college has 10 basketball players. A 5 member team and a captain will be selected out of these 10 players. How many different selections can be made?
We can select the 5 member team out of the 10 in 10C5 ways = 252 ways. The captain can be selected from amongst the remaining 5 players in 5 ways. Therefore, total ways the selection of 5 players and a captain can be made = 252*5 = 1260.
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In how many ways can 5 different toys be packed in 3 identical boxes such that no. box is empty, if any of the boxes may hold all of the toys?
Number of ways of achieving the second option 3 - 1 - 1
Three toys out of the 5 can be selected in 5C3ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.
Therefore, total number of ways of getting the 3 - 1 - 1 option is 5C3 = 10 = 10 ways.
Total ways in which the 5 toys can be packed in 3 identical boxes
= number of ways of achieving Case a + number of ways of achieving Case b
= 15 + 10 = 25 ways.
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How many 4-bit digit numbers that do not contain the digits 3 or 6 are there?
Left digit can be any of the (1,2,4,5,7,8,9) --> 7 ways
Other digits can be (0,1,2,4,5,7,8,9) --> 8 ways
So the total ways = 7*8*8*8 = 3584 or 3584 numbers without any 3 or 6.
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A tourist wants to visit 3 or more of the 5 major cities in India: Chennai, Bangalore, Mumbai, Delhi and Kolkata. In how many ways can be plan his tour such that Chennai is always includes? Two plan of the tour are different if the cities in the tour or the order of the cities are different.
As chennai is always there,
So tourist have to choose 2 cities out of 4 cities(excluding chennai) and this can be done by
Chennai must always be includes.
for 3 cities 4C2*3! => 6*3! => 36
for 4 cities 4C3*4! => 4*4! => 96
for 5 cities 4C4*5! => 1*5! => 120
So, Total = 5!+4*4!+6*3!
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Consider all permutation (i.e. arrangements) of the digits 1 2 and 3. We will say that a hit has been scored if at least one digit occurs in its proper position in the permutation i.e. if one occurs in the 1st position or 2 in the 2nd position or 3 in the 3rd position. In how many of these permutations is a hit scored?
1 can be arrange in 123, 132
2 can be arrange in 321
3 can be arrange in 213
So, No. of possobolity = 4.
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A 3*3 grid is colored using red and blue colors, such that if we rotate the grid about its centre in the plane by 180 degrees, the grid looks the same. The number of ways to color the grid this way is:
Oranges can be packed in sets of 10 oranges in box type A or 25 orange in box type B. A carton comprising of 1000 oranges of type A and type B boxes is packed. How many different combinations are possible in the number of type A and type B boxes while organizing the oranges?
The expression will be 10x+25y=1000 ----------(1)
The exp (1) satisfied only with multiples of 2 up to 38.
so no. of combinations are 19.
if oranges are packed in type A boxes only= 1 way
if oranges are packed in type B boxes only= 1 way
Hence total = 19+1+1 = 21
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A person starts writing all 4 digits numbers.How many times had he written the digit 2?
When 2 is at unit place=9*10*10*1
When 2 is at tenth place=9*10*1*10
When 2 is at hundred place=9*1*10*10
When 2 is at thousand place = 1*10*10*10
So Total no. = 900 + 900 + 900 + 1000 = 3700
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