Chapter 5: Problem 391
The number of zeros at the end of \(100 !\) is (a) 20 (b) 22 (c) 24 (d) 26
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Chapter 5: Problem 391
The number of zeros at the end of \(100 !\) is (a) 20 (b) 22 (c) 24 (d) 26
These are the key concepts you need to understand to accurately answer the question.
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In a certain test there are n questions. In this test \(2^{\mathrm{k}}\) students gave wrong answers to at least \((\mathrm{n}-\mathrm{k})\) question. \(\mathrm{k}=0,1,2 \ldots \mathrm{n} .\) If the no. of wrong answers is 4095 then value of \(\mathrm{n}\) is (a) 11 (b) 12 (c) 13 (d) 15
If \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}\) and \({ }^{\mathrm{n}} \mathrm{P}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{P}_{\mathrm{r}+1}\), then the value of \(\mathrm{n}\) is (a) 3 (b) 4 (c) 2 (d) 5
Three boys and three girls are to be seated around a round table in a circle. Among them the boy \(\mathrm{X}\) does not want any girl neighbour and the girl \(\mathrm{Y}\) does not want any boy neighbour then the no. of arrangement is (a) 2 (b) 4 (c) 23 (d) 33
8 chairs are numbered 1 to 8 Two women and 3 men wish to occupy one chair each first the women choose the chairs marked, 1 to 4 then the men select from among the remaining chairs. The number of seating arrangements is (a) 1250 (b) 1360 (c) 1440 (d) 1560
The number of straight lines that can be drawn out of 10 points of which 7 are collinear is (a) 22 (b) 23 (c) 24 (d) 25
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