Permutation and Combinations Quiz 1

We know that the value of log y to the base x is 1
when x and y value will same.
So the answer to the question is 10(c).
If the answer is not right send the correct one.

Only one employee leave request for a particular day

five employees requested

so one\'s probability => 1/5(and it is same for all)

at least two means -> 2 or 3 or 4 or 5 or 6 or 7

so option b (b) 7C2 +7 C3 +7 C4 + 7C5 +7 C6 + 7C7

A column compared with -> B , C , D , E , F -> 5 comparisons

B column compared with -> C , D , E , F -> 4 comparisons

c column compared with -> D , E , F -> 3 comparisons

D column compared with -> E , F -> 2 comparisons

A column compared with -> F -> 1 comparisons

Now all pair comparison finished, no pair left to be compared; so total comparisons = 15

NOTE here it is not said that each of the five letters should be unique or no repetition allowed

so we will assume that repetition allowed. Total letters: 26 consonants: 21

26C1 * 26C1 * 26C1 * 26C1 * 21C1

=> 26^4 * 21

There is a total of 26 black cards and 26 red cards in a pack.
case1: black black black black
case2: black black black red
case3: black red black black
case4: black red black red
We have to choose a black card on the first draw out of 26 black cards. After the first draw, there will be 25 black cards remaining. If the second draw will be black, we have to choose it from 25 remainings. Then in the third draw, we will have 24 black cards and we have to choose from those. For the fourth draw, we will have 52-3=49 cards remaining since we have already chosen 3 cards for three draws.
Now, if the second draw is red then we have to choose it from 26 red cards. we will have to choose black cards for the first draw from 26 black cards and second draw from remaining 25 black cards. In this case, also we have to choose the fourth draw from 52-3=49 cards since we have already chosen 3 cards for three draws.

So, the number of ways we can do this is 26*25*24*49 + 26*26*25*49 = 1,592,500, so option 4 is correct.

We know, Selecting 1 book out of p books is Pc1 = P. Selecting no book out of P books is also considered as a count. So a number of ways we can select for 1 subject are Pc1(selecting 1 book of that type) + Pc0(not selecting that book). = P+1.
So for n books = (P+1))(P+1)(P+1)... n times.. = (P+1)^n.
However, in all the cases we are counting the ways where we are not selecting any book(i.e Pc0).
So there will be only one way in which we are not selecting any book from any category
Hence the answer is ((P+1)^n)-1.

216
combinations: 0+1+2+4+5 and 1+2+3+4+5 are the only possible ways here (because there sum is divisible by 3) so,
(5!-4!)+5!

If ram draws 1 then Sam has 22 choices. If ram draws 2 then Sam has 21 choices and so on..

So, the probability that Sam has drawn a card greater than Ram is

= 1/23*22/23 + 1/23*21/23 + 1/23*20/23 + .... + 1/23*1/23

=1/23*[(22+21+...+10)/23]

=> 1/23*{22/2*[1+22]}/23

= 252/529 or 84/143 ans

Combinations from A and from B:: 4/6 ; 5/5 ; 6/4
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# of ways to get the 4/6 pattern: 6C4*7C6 = 15*7 = 105
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# of ways to get the 5/5 pattern: 6C5*7C5 = 6*21 = 126
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# of ways to get the 6/4 pattern: 6C6*7C4 = 1*35 = 35
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Total # of ways: 266